The main features of the E-book are:- i. The E-book acts simultaneously as a guide book. ii. All the fundamental concepts have been inserted. iii. A fairly large number of examples have been worked out properly. iv. Exercises are graded and fully solved. v. The E-book will serve as a valuable guide for various competitive Examinations.

1. Rational Fractions
{𝑀𝑒𝑡ℎ𝑜𝑑 𝑜𝑓 𝑏𝑟𝑒𝑎𝑘𝑖𝑛𝑔 𝑢𝑝 𝑖𝑛𝑡𝑜 𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛𝑠}
Factor in the denominator Corresponding partial fraction
(i) (𝑥 − 𝑎)
(ii) (𝑥 − 𝑏)2
(iii) (𝑥 − 𝑐)3
(iv) (𝑎𝑥2
+ 𝑏𝑥 + 𝑐)
(i)
𝐴
(𝑥−𝑎)
(ii)
𝐴
(𝑥−𝑏)
+
𝐵
(𝑥−𝑏)2
(iii)
𝐴
(𝑥−𝑐)
+
𝐵
(𝑥−𝑐)2 +
𝐶
(𝑥−𝑐)3
(iv)
𝐴𝑥+𝐵
(𝑎𝑥2+ 𝑏𝑥+𝑐)
Exercise .A
Integrate the following:
1. ∫
(𝑥−1)𝑑𝑥
(𝑥−2)(𝑥−3)
Solution: Let 𝐼 = ∫
(𝑥−1)𝑑𝑥
(𝑥−2)(𝑥−3)
Let
(𝑥−1)
(𝑥−2)(𝑥−3)
=
𝐴
(𝑥−2)
+
𝐵
(𝑥−3)
---------- (a)
(𝑥−1)
(𝑥−2)(𝑥−3)
=
𝐴(𝑥−3)+ 𝐵(𝑥−2)
(𝑥−2)(𝑥−3)
 (𝑥 − 1) = 𝐴(𝑥 − 3) + 𝐵(𝑥 − 2) -------- (b)
Putting 𝑥 = 2 and 𝑥 = 3 in equation (b), we get
𝐴 = −1 and 𝐵 = 2
Now,
𝐼 = ∫
(𝑥−1)𝑑𝑥
(𝑥−2)(𝑥−3)
𝐼 = ∫ {
𝐴
(𝑥−2)
+
𝐵
(𝑥−3)
} 𝑑𝑥 Using (a)
𝐼 = ∫
𝐴
(𝑥−2)
𝑑𝑥 + ∫
𝐵
(𝑥−3)
𝑑𝑥

2. 𝐼 = 𝐴 ∫
𝑑𝑥
(𝑥−2)
+ 𝐵 ∫
𝑑𝑥
(𝑥−3)
𝐼 = 𝐴 log(𝑥 − 2) + 𝐵 log(𝑥 − 3)
𝐼 = −log(𝑥 − 2) + 2 log(𝑥 − 3)
𝐼 = 2 log(𝑥 − 3) − log(𝑥 − 2)
2. ∫
𝑥 𝑑𝑥
(𝑥−𝑎)(𝑥−𝑏)
Solution: Let 𝐼 = ∫
𝑥 𝑑𝑥
(𝑥−𝑎)(𝑥−𝑏)
Let,
𝑥
(𝑥−𝑎)(𝑥−𝑏)
=
𝐴
(𝑥−𝑎)
+
𝐵
(𝑥−𝑏)
---------- (a)
𝑥
(𝑥−𝑎)(𝑥−𝑏)
=
𝐴(𝑥−𝑏)+ 𝐵 (𝑥−𝑎)
(𝑥−𝑎) (𝑥−𝑏)
𝑥 = 𝐴(𝑥 − 𝑏) + 𝐵 (𝑥 − 𝑎) ----------- (b)
Putting 𝑥 = 𝑎 and 𝑥 = 𝑏 in (b), we get
𝐴 =
𝑎
(𝑎−𝑏)
and 𝐵 =
𝑏
(𝑏−𝑎)
𝑁𝑜𝑤,
𝐼 = ∫
𝑥 𝑑𝑥
(𝑥−𝑎)(𝑥−𝑏)
𝐼 = ∫ {
𝐴
(𝑥−𝑎)
+
𝐵
(𝑥−𝑏)
} 𝑑𝑥 Using (a)
𝐼 = ∫
𝐴
(𝑥−𝑎)
𝑑𝑥 + ∫
𝐵
(𝑥−𝑏)
𝑑𝑥
𝐼 = 𝐴 ∫
𝑑𝑥
(𝑥−𝑎)
+ 𝐵 ∫
𝑑𝑥
(𝑥−𝑏)
𝐼 =
𝑎
(𝑎−𝑏)
∫
𝑑𝑥
(𝑥−𝑎)
+
𝑏
(𝑏−𝑎)
∫
𝑑𝑥
(𝑥−𝑏)
𝐼 =
𝑎
(𝑎−𝑏)
log(𝑥 − 𝑎) +
𝑏
−(𝑎−𝑏)
log(𝑥 − 𝑏)
𝐼 =
1
(𝑎−𝑏)
(𝑎 log(𝑥 − 𝑎) − 𝑏 log(𝑥 − 𝑏))
3. ∫
(𝑥−1) 𝑑𝑥
(𝑥+2)(𝑥−3)
Solution: Let 𝐼 = ∫
(𝑥−1) 𝑑𝑥
(𝑥+2)(𝑥−3)

3. Let,
(𝑥−1)
(𝑥+2)(𝑥−3)
=
𝐴
(𝑥+2)
+
𝐵
(𝑥−3)
-------- (a)
(𝑥−1)
(𝑥+2)(𝑥−3)
=
𝐴(𝑥−3)+ 𝐵(𝑥+2)
(𝑥+2)(𝑥−3)
(𝑥 − 1) = 𝐴(𝑥 − 3) + 𝐵(𝑥 + 2) --------- (b)
Putting 𝑥 = 3 and 𝑥 = −2 in equation (b), we get
𝐴 =
3
5
and 𝐵 =
2
5
Now,
𝐼 = ∫
(𝑥−1) 𝑑𝑥
(𝑥+2)(𝑥−3)
𝐼 = ∫ (
𝐴
(𝑥+2)
+
𝐵
(𝑥−3)
) 𝑑𝑥 Using (a)
𝐼 = ∫
𝐴
(𝑥+2)
𝑑𝑥 + ∫
𝐵
(𝑥−3)
𝑑𝑥
𝐼 = 𝐴 ∫
𝑑𝑥
(𝑥+2)
+ 𝐵 ∫
𝑑𝑥
(𝑥−3)
𝐼 =
3
5
∫
𝑑𝑥
(𝑥+2)
+
2
5
∫
𝑑𝑥
(𝑥−3)
𝐼 =
3
5
log(𝑥 + 2) +
2
5
log(𝑥 − 3)
𝐼 =
1
5
(3 log(𝑥 + 2) + 2 log(𝑥 − 3))
4. (i) ∫
𝑥 𝑑𝑥
𝑥2−12𝑥+35
(ii) ∫
3𝑥 𝑑𝑥
𝑥2− 𝑥−2
(i) Solution: Let 𝐼 = ∫
𝑑𝑥
𝑥2−12𝑥+35
𝐼 = ∫
𝑥 𝑑𝑥
𝑥2−7𝑥−5𝑥+35
𝐼 = ∫
𝑥 𝑑𝑥
𝑥(𝑥−7)−5(𝑥−7)
𝐼 = ∫
𝑥 𝑑𝑥
(𝑥−5)(𝑥−7)
Let,
𝑥
(𝑥−5)(𝑥−7)
=
𝐴
(𝑥−5)
+
𝐵
(𝑥−7)
-------- (a)
𝑥
(𝑥−5)(𝑥−7)
=
𝐴 (𝑥−7)+ 𝐵 (𝑥−5)
(𝑥−5)
𝑥 = 𝐴 (𝑥 − 7) + 𝐵 (𝑥 − 5) ------------ (b)

4. Putting 𝑥 = 7 and 𝑥 = 5 in equation (b), we get
𝐴 = −
5
2
and 𝐵 =
7
2
Now,
𝐼 = ∫
𝑥 𝑑𝑥
(𝑥−5)(𝑥−7)
𝐼 = ∫ (
𝐴
(𝑥−5)
+
𝐵
(𝑥−7)
) 𝑑𝑥 Using (a)
𝐼 = ∫
𝐴
(𝑥−5)
𝑑𝑥 + ∫
𝐵
(𝑥−7)
𝑑𝑥
𝐼 = 𝐴 ∫
𝑑𝑥
(𝑥−5)
+ 𝐵 ∫
𝑑𝑥
(𝑥−7)
𝐼 = −
5
2
∫
𝑑𝑥
(𝑥−5)
+
7
2
∫
𝑑𝑥
(𝑥−7)
𝐼 = −
5
2
log(𝑥 − 5) +
7
2
log(𝑥 − 7)
𝐼 =
1
2
(7 log(𝑥 − 7) −5log(𝑥 − 5))
(ii) Solution: Let 𝐼 = ∫
3𝑥 𝑑𝑥
𝑥2− 𝑥−2
𝐼 = ∫
3𝑥 𝑑𝑥
𝑥2−2 𝑥+𝑥−2
𝐼 = ∫
3𝑥 𝑑𝑥
𝑥(𝑥−2)+1(𝑥−2)
𝐼 = ∫
3𝑥 𝑑𝑥
(𝑥+1)(𝑥−2)
Let,
3𝑥
(𝑥+1)(𝑥−2)
=
𝐴
(𝑥+1)
+
𝐵
(𝑥−2)
------ (a)
3𝑥
(𝑥+1)(𝑥−2)
=
𝐴(𝑥−2)+ 𝐵(𝑥+1)
(𝑥+1)(𝑥−2)
3𝑥 = 𝐴(𝑥 − 2) + 𝐵(𝑥 + 1) -------- (b)
Putting 𝑥 = 2 and 𝑥 = −1, we get
𝐴 = 1 and 𝐵 = 2
Now,
𝐼 = ∫
3𝑥 𝑑𝑥
(𝑥+1)(𝑥−2)
𝐼 = ∫ (
𝐴
(𝑥+1)
+
𝐵
(𝑥−2)
) 𝑑𝑥

5. 𝐼 = ∫
𝐴
(𝑥+1)
𝑑𝑥 + ∫
𝐵
(𝑥−2)
𝑑𝑥
𝐼 = 𝐴 ∫
𝑑𝑥
(𝑥+1)
+ 𝐵 ∫
𝑑𝑥
(𝑥−2)
𝐼 = ∫
𝑑𝑥
(𝑥+1)
+ 2 ∫
𝑑𝑥
(𝑥−2)
𝐼 = log(𝑥 + 1) + 2 log(𝑥 − 2)
𝐼 = log(𝑥 + 1) + log(𝑥 − 2)2
𝐼 = log{(𝑥 − 2)2(𝑥 + 1)}
5. ∫
𝑥2 𝑑𝑥
(𝑥−1)(𝑥−2)(𝑥−3)
Solution: Let 𝐼 = ∫
𝑥2 𝑑𝑥
(𝑥−1)(𝑥−2)(𝑥−3)
Let,
𝑥2
(𝑥−1)(𝑥−2)(𝑥−3)
=
𝐴
(𝑥−1)
+
𝐵
(𝑥−2)
+
𝐶
(𝑥−3)
-------- (a)
𝑥2
(𝑥−1)(𝑥−2)(𝑥−3)
=
𝐴(𝑥−2)(𝑥−3)+ 𝐵(𝑥−1)(𝑥−3) + 𝐶(𝑥−1)(𝑥−2)
(𝑥−1)(𝑥−2)(𝑥−3)
𝑥2
= 𝐴(𝑥 − 2)(𝑥 − 3) + 𝐵(𝑥 − 1)(𝑥 − 3) + 𝐶(𝑥 − 1)(𝑥 − 2) -------- (b)
Putting 𝑥 = 1 , 𝑥 = 2 and 𝑥 = 3 in equation (b), we get
𝐴 =
1
2
, 𝐵 = −4 and 𝐶 =
9
2
Now,
𝐼 = ∫
𝑥2 𝑑𝑥
(𝑥−1)(𝑥−2)(𝑥−3)
𝐼 = ∫ (
𝐴
(𝑥−1)
+
𝐵
(𝑥−2)
+
𝐶
(𝑥−3)
) 𝑑𝑥
𝐼 = ∫
𝐴
(𝑥−1)
𝑑𝑥 + ∫
𝐵
(𝑥−2)
𝑑𝑥 + ∫
𝐶
(𝑥−3)
𝑑𝑥
𝐼 = 𝐴 ∫
𝑑𝑥
(𝑥−1)
+ 𝐵 ∫
𝑑𝑥
(𝑥−2)
+ 𝐶 ∫
𝑑𝑥
(𝑥−3)
𝐼 = 𝐴 log(𝑥 − 1) + 𝐵 log(𝑥 − 2) + 𝐶 log(𝑥 − 3)
𝐼 =
1
2
log(𝑥 − 1) − 4 log(𝑥 − 2) +
9
2
log(𝑥 − 3)
𝑛 log 𝑥 = log 𝑥𝑛
log 𝐴 + log 𝐵 = log 𝐴𝐵

6. 6. (i) ∫
(2𝑥+3) 𝑑𝑥
𝑥3+𝑥2− 2𝑥
(ii) ∫
(𝑥2+1)𝑑𝑥
𝑥(𝑥2−1)
(i) Solution: Let 𝐼 = ∫
(2𝑥+3) 𝑑𝑥
𝑥3+𝑥2− 2𝑥
𝐼 = ∫
(2𝑥+3) 𝑑𝑥
𝑥(𝑥2+𝑥−2)
𝐼 = ∫
(2𝑥+3) 𝑑𝑥
𝑥(𝑥2+2𝑥−𝑥−2)
𝐼 = ∫
(2𝑥+3) 𝑑𝑥
𝑥{𝑥(𝑥+2)−1(𝑥+2)}
𝐼 = ∫
(2𝑥+3) 𝑑𝑥
𝑥{(𝑥−1)(𝑥+2)}
𝐼 = ∫
(2𝑥+3) 𝑑𝑥
𝑥 (𝑥−1)(𝑥+2)
Let,
(2𝑥+3)
𝑥 (𝑥−1)(𝑥+2)
=
𝐴
𝑥
+
𝐵
(𝑥−1)
+
𝐶
(𝑥+2)
------- (a)
(2𝑥+3)
𝑥 (𝑥−1)(𝑥+2)
=
𝐴 (𝑥−1)(𝑥+2)+ 𝐵 𝑥 (𝑥+2)+ 𝐶 𝑥 (𝑥−1)
𝑥 (𝑥−1)(𝑥+2)
(2𝑥 + 3) = 𝐴 (𝑥 − 1)(𝑥 + 2) + 𝐵 𝑥 (𝑥 + 2) + 𝐶 𝑥 (𝑥 − 1) ----- (b)
Putting, = 1 , 𝑥 = −2 and 𝑥 = 0 in equation (b), we get
𝐴 = −
3
2
, 𝐵 =
5
3
and 𝐶 = −
1
6
𝑁𝑜𝑤,
𝐼 = ∫
(2𝑥+3) 𝑑𝑥
𝑥 (𝑥−1)(𝑥+2)
𝐼 = ∫ {
𝐴
𝑥
+
𝐵
(𝑥−1)
+
𝐶
(𝑥+2)
} 𝑑𝑥
𝐼 = ∫
𝐴
𝑥
𝑑𝑥 + ∫
𝐵
(𝑥−1)
𝑑𝑥 + ∫
𝐶
(𝑥+2)
𝑑𝑥
𝐼 = 𝐴 ∫
𝑑𝑥
𝑥
+ 𝐵 ∫
𝑑𝑥
(𝑥−1)
+ 𝐶 ∫
𝑑𝑥
(𝑥+2)
𝐼 = 𝐴 log 𝑥 + 𝐵 log(𝑥 − 1) + 𝐶 log(𝑥 + 2)
𝐼 = −
3
2
log 𝑥 +
5
3
log(𝑥 − 1) −
1
6
log(𝑥 + 2)
𝐼 =
5
3
log(𝑥 − 1) −
3
2
log 𝑥 −
1
6
log(𝑥 + 2)

7. (ii) Solution: Let 𝐼 = ∫
(𝑥2+1)𝑑𝑥
𝑥(𝑥2−1)
𝐼 = ∫
(𝑥2+1)𝑑𝑥
𝑥{(𝑥+1)(𝑥−1)}
𝐼 = ∫
(𝑥2+1)𝑑𝑥
𝑥 (𝑥+1)(𝑥−1)
Let,
(𝑥2+1)
𝑥 (𝑥+1)(𝑥−1)
=
𝐴
𝑥
+
𝐵
(𝑥+1)
+
𝐶
(𝑥−1)
----------- (a)
(𝑥2+1)
𝑥 (𝑥+1)(𝑥−1)
=
𝐴 (𝑥+1)(𝑥−1)+ 𝐵 𝑥 (𝑥−1)+ 𝐶 𝑥 (𝑥+1)
𝑥 (𝑥+1)(𝑥−1)
(𝑥2
+ 1) = 𝐴 (𝑥 + 1)(𝑥 − 1) + 𝐵 𝑥 (𝑥 − 1) + 𝐶 𝑥 (𝑥 + 1) ---------- (b)
Putting, 𝑥 = 0, 𝑥 = 1 and 𝑥 = −1 in equation (b), we get
𝐴 = −1 , 𝐵 = 1 and 𝐶 = 1
Now,
𝐼 = ∫
(𝑥2+1)𝑑𝑥
𝑥 (𝑥+1)(𝑥−1)
𝐼 = ∫ {
𝐴
𝑥
+
𝐵
(𝑥+1)
+
𝐶
(𝑥−1)
} 𝑑𝑥
𝐼 = ∫
𝐴
𝑥
𝑑𝑥 + ∫
𝐵
(𝑥+1)
𝑑𝑥 + ∫
𝐶
(𝑥−1)
𝑑𝑥
𝐼 = 𝐴 ∫
𝑑𝑥
𝑥
+ 𝐵 ∫
𝑑𝑥
(𝑥+1)
+ 𝐶 ∫
𝑑𝑥
(𝑥−1)
𝐼 = 𝐴 log 𝑥 + 𝐵 log(𝑥 + 1) + 𝐶 log(𝑥 − 1)
𝐼 = −log 𝑥 + log(𝑥 + 1) + log(𝑥 − 1)
𝐼 = −log 𝑥 + log{(𝑥 + 1)(𝑥 − 1)}
𝐼 = −log 𝑥 + log{(𝑥2
− 1)}
𝐼 = −log 𝑥 + log(𝑥2
− 1)
𝐼 = log(𝑥2
− 1) − log 𝑥
7. (i) ∫
𝑥3 𝑑𝑥
𝑥2+ 7𝑥+12
(ii) ∫
(𝑥−1)(𝑥−5)
(𝑥−2)(𝑥−4)
𝑑𝑥
log 𝐴 + log 𝐵 = log 𝐴𝐵

8. (i) Solution: Let 𝐼 = ∫
𝑥3 𝑑𝑥
𝑥2+ 7𝑥+12
𝐼 = ∫
𝑥3
𝑥2+ 7𝑥 + 12
𝑑𝑥
𝐼 = ∫ (𝑥 − 7 +
37𝑥+84
𝑥2+ 7𝑥 + 12
) 𝑑𝑥
𝐼 = ∫ 𝑥 𝑑𝑥 − 7 ∫ 𝑑𝑥 + ∫
37𝑥+84
𝑥2+ 7𝑥 + 12
𝑑𝑥
𝐼 =
1
2
𝑥2
− 7𝑥 + ∫
37𝑥+84
𝑥2+ 7𝑥 + 12
𝑑𝑥
𝐼 =
1
2
𝑥2
− 7𝑥 + ∫
37𝑥+84
𝑥2+ 3𝑥+4𝑥+12
𝑑𝑥
𝐼 =
1
2
𝑥2
− 7𝑥 + ∫
37𝑥+84
(𝑥+3)(𝑥+4)
𝑑𝑥
Let,
37𝑥+84
(𝑥+3)(𝑥+4)
=
𝐴
(𝑥+3)
+
𝐵
(𝑥+4)
----------- (a)
37𝑥+84
(𝑥+3)(𝑥+4)
=
𝐴(𝑥+4)+ 𝐵 (𝑥+3)
(𝑥+3) (𝑥+4)
37𝑥 + 84 = 𝐴(𝑥 + 4) + 𝐵 (𝑥 + 3) ---------- (b)
𝑃𝑢𝑡𝑡𝑖𝑛𝑔, 𝑥 = −3 and 𝑥 = −4 in equation (b), we get
𝐴 = −27 and 𝐵 = 64
Now,
𝐼 =
1
2
𝑥2
− 7𝑥 + ∫
37𝑥+84
(𝑥+3)(𝑥+4)
𝑑𝑥
𝐼 =
1
2
𝑥2
− 7𝑥 + ∫ (
𝐴
(𝑥+3)
+
𝐵
(𝑥+4)
) 𝑑𝑥
𝐼 =
1
2
𝑥2
− 7𝑥 + ∫
𝐴
(𝑥+3)
𝑑𝑥 + ∫
𝐵
(𝑥+4)
𝑑𝑥
𝐼 =
1
2
𝑥2
− 7𝑥 + 𝐴 ∫
𝑑𝑥
(𝑥+3)
+ 𝐵 ∫
𝑑𝑥
(𝑥+4)
𝐼 =
1
2
𝑥2
− 7𝑥 + 𝐴 log(𝑥 + 3) + 𝐵 log(𝑥 + 4)
𝐼 =
1
2
𝑥2
− 7𝑥 − 27 log(𝑥 + 3) + 64 log(𝑥 + 4)
(ii) Solution: Let 𝐼 = ∫
(𝑥−1)(𝑥−5)
(𝑥−2)(𝑥−4)
𝑑𝑥
𝐼 = ∫
𝑥2−6𝑥+5
𝑥2−6𝑥+8
𝑑𝑥
𝑥2
+ 7𝑥 + 12 ⟌𝑥3
( 𝑥 − 7
𝑥3
+ 7𝑥2
+ 12𝑥
_____________________
−7𝑥2
− 12𝑥
−7𝑥2
− 49𝑥 − 84
_________________________
37𝑥 + 84

9. 𝐼 = ∫ (
𝑥2−6𝑥+8
𝑥2−6𝑥+8
−
3
𝑥2−6𝑥+8
) 𝑑𝑥
𝐼 = ∫ (1 −
3
𝑥2−6𝑥+8
) 𝑑𝑥
𝐼 = ∫ 1 𝑑𝑥 − ∫
3
𝑥2−6𝑥+8
𝑑𝑥
𝐼 = 𝑥 − 3 ∫
1
𝑥2−6𝑥+8
𝑑𝑥
𝐼 = 𝑥 − 3 ∫
1
(𝑥−2)(𝑥−4)
𝑑𝑥
Take,
1
(𝑥−2)(𝑥−4)
=
𝐴
(𝑥−2)
+
𝐵
(𝑥−4)
----------- (a)
1
(𝑥−2)(𝑥−4)
=
𝐴 (𝑥−4) +𝐵 (𝑥−2)
(𝑥−2)
1 = 𝐴 (𝑥 − 4) + 𝐵 (𝑥 − 2) ------------- (b)
Putting, 𝑥 = 2 and 𝑥 = 4, we get
𝐴 = −
1
2
and 𝐵 =
1
2
Now,
𝐼 = 𝑥 − 3 ∫
1
(𝑥−2)(𝑥−4)
𝑑𝑥
𝐼 = 𝑥 − 3 ∫ {
𝐴
(𝑥−2)
+
𝐵
(𝑥−4)
} 𝑑𝑥 Using (a)
𝐼 = 𝑥 − 3 {∫
𝐴
(𝑥−2)
𝑑𝑥 + ∫
𝐵
(𝑥−4)
𝑑𝑥}
𝐼 = 𝑥 − 3 {𝐴∫
𝑑𝑥
(𝑥−2)
+ 𝐵 ∫
𝑑𝑥
(𝑥−4)
}
𝐼 = 𝑥 − 3{𝐴 log(𝑥 + 2) + 𝐵 log(𝑥 − 4)}
𝐼 = 𝑥 − 3 {−
1
2
log(𝑥 + 2) +
1
2
log(𝑥 − 4)}
𝐼 = 𝑥 −
3
2
{−log(𝑥 + 2) + log(𝑥 − 4)}
𝐼 = 𝑥 −
3
2
{log(𝑥 − 4) − log(𝑥 − 2)}
𝐼 = 𝑥 −
3
2
log
𝑥−4
𝑥−2
8. (i) ∫
1−3𝑥2
3𝑥−𝑥3 𝑑𝑥 (ii) ∫
𝑥 𝑑𝑥
(3−𝑥)(3+2𝑥)
log 𝐴 − log 𝐵 = log
𝐴
𝐵

10. (i) Solution: Let 𝐼 = ∫
1−3𝑥2
3𝑥−𝑥3 𝑑𝑥
𝐼 = ∫
1
3𝑥−𝑥3 𝑑𝑥 − ∫
3𝑥2
3𝑥−𝑥3 𝑑𝑥
𝐼 = ∫
1
𝑥(3−𝑥2)
𝑑𝑥 − ∫
3𝑥2
𝑥(3−𝑥2)
𝑑𝑥
𝐼 = ∫
1
𝑥(√3 +𝑥)(√𝑥−3)
𝑑𝑥 − 3 ∫
𝑥
3−𝑥2 𝑑𝑥
𝐼 = 𝐼1 − 3𝐼2 Where, 𝐼1 = ∫
1
𝑥(√3 +𝑥)(√𝑥−3)
𝑑𝑥 and 𝐼2 = ∫
𝑥
3−𝑥2 𝑑𝑥
Here,
𝐼1 = ∫
1
𝑥(√3 +𝑥)(√3−𝑥)
𝑑𝑥
Let,
1
𝑥(√3 +𝑥)(√3−𝑥)
=
𝐴
𝑥
+
𝐵
(√3 +𝑥)
+
𝐶
(√3−𝑥)
------------ (b)
1
𝑥(√3 +𝑥)(√𝑥−3)
=
𝐴(√3 +𝑥)(√3−𝑥)+ 𝐵 𝑥(√𝑥−3)+ 𝐶 𝑥(√3 +𝑥)
𝑥(√3 +𝑥)(3−𝑥)
1 = 𝐴(√3 + 𝑥)(√3 − 𝑥) + 𝐵 𝑥(√3 − 𝑥) + 𝐶 𝑥(√3 + 𝑥) ------------ (b)
Putting, 𝑥 = 0 , 𝑥 = √3 and 𝑥 = −√3 in equation (b) , we get
𝐴 =
1
3
, 𝐵 = −
1
6
and 𝐶 =
1
6
Now,
𝐼1 = ∫
1
𝑥(√3 +𝑥)(√3−𝑥)
𝑑𝑥
𝐼1 = ∫ {
𝐴
𝑥
+
𝐵
(√3 +𝑥)
+
𝐶
(√3−𝑥)
} 𝑑𝑥
𝐼1 = ∫
𝐴
𝑥
𝑑𝑥 + ∫
𝐵
(√3 +𝑥)
𝑑𝑥 + ∫
𝐶
(√3−𝑥)
𝑑𝑥
𝐼1 = 𝐴 ∫
𝑑𝑥
𝑥
+ 𝐵 ∫
𝑑𝑥
(√3 +𝑥)
+ 𝐶 ∫
𝑑𝑥
(√3−𝑥)
𝐼1 = 𝐴 log 𝑥 + 𝐵 log(√3 + 𝑥) + 𝐶 ∫
𝑑𝑥
(√3−𝑥)
𝐼1 = 𝐴 log 𝑥 + 𝐵 log(√3 + 𝑥) + 𝐶 ∫
𝑑𝑥
(√3−𝑥)
𝐼1 = 𝐴 log 𝑥 + 𝐵 log(√3 + 𝑥) − 𝐶 log(√3 − 𝑥)
𝐼1 =
1
3
log 𝑥 −
1
6
log(√3 + 𝑥) −
1
6
log(√3 − 𝑥)
𝐼1 =
1
3
log 𝑥 −
1
6
[log(√3 + 𝑥) + log(√3 − 𝑥)]

11. 𝐼1 =
1
3
log 𝑥 −
1
6
[log{(√3 + 𝑥)(√3 − 𝑥)}]
𝐼1 =
1
3
log 𝑥 −
1
6
[log{(3 − 𝑥2)}]
𝐼1 =
1
3
log 𝑥 −
1
6
log{(3 − 𝑥2)}
𝐼1 =
1
3
log 𝑥 −
1
6
log(3 − 𝑥2)
And
𝐼2 = ∫
𝑥
3−𝑥2 𝑑𝑥
Taking, 3 − 𝑥2
= 𝑢  𝑥 𝑑𝑥 = −
1
2
𝑑𝑢
𝐼2 = ∫
−
1
2
𝑑𝑢
𝑢
𝐼2 = −
1
2
∫
𝑑𝑢
𝑢
𝐼2 = −
1
2
log 𝑢
𝐼2 = −
1
2
log(3 − 𝑥2)
𝑇ℎ𝑢𝑠,
𝐼 = 𝐼1 − 3𝐼2
𝐼 =
1
3
log 𝑥 −
1
6
log(3 − 𝑥2) − 3 (−
1
2
log(3 − 𝑥2))
𝐼 =
1
3
log 𝑥 −
1
6
log(3 − 𝑥2) +
3
2
log(3 − 𝑥2)
𝐼 =
1
3
log 𝑥 +
4
3
log(3 − 𝑥2)
𝐼 =
1
3
{log 𝑥 + 4 log(3 − 𝑥2)}
𝐼 =
1
3
{log 𝑥 + log(3 − 𝑥2)4}
𝐼 =
1
3
log{𝑥(3 − 𝑥2)4}
(ii) Solution: Let 𝐼 = ∫
𝑥 𝑑𝑥
(3−𝑥)(3+2𝑥)
Let,
𝑥
(3−𝑥)(3+2𝑥)
=
𝐴
(3−𝑥)
+
𝐵
(3+2𝑥)
-------- (a)
𝑥
(3−𝑥)(3+2𝑥)
=
𝐴(3+2𝑥)+ 𝐵(3−𝑥)
(3−𝑥)(3+2𝑥)
𝑛 log 𝑥 = log 𝑥𝑛
log 𝐴 + log 𝐵 = log 𝐴𝐵

12. 𝑥 = 𝐴(3 + 2𝑥) + 𝐵(3 − 𝑥) ------------ (b)
P𝑢𝑡𝑡𝑖𝑛𝑔 𝑥 = 3 and 𝑥 = −
3
2
in equation (b), we get
𝐴 =
1
3
and 𝐵 = −
1
3
Now,
𝐼 = ∫
𝑥 𝑑𝑥
(3−𝑥)(3+2𝑥)
𝐼 = ∫ {
𝐴
(3−𝑥)
+
𝐵
(3+2𝑥)
} 𝑑𝑥
𝐼 = ∫
𝐴
(3−𝑥)
𝑑𝑥 + ∫
𝐵
(3+2𝑥)
𝑑𝑥
𝐼 = 𝐴 ∫
𝑑𝑥
(3−𝑥)
+ 𝐵 ∫
𝑑𝑥
(3+2𝑥)
𝑇𝑎𝑘𝑒, 3 − 𝑥 = 𝑢  𝑑𝑥 = −𝑑𝑢 and 3 + 2𝑥 = 𝑣  𝑑𝑥 =
1
2
𝑑𝑣
𝐼 = 𝐴 ∫
− 𝑑𝑢
𝑢
+ 𝐵 ∫
1
2
𝑑𝑣
𝑣
𝐼 = −𝐴 ∫
𝑑𝑢
𝑢
+
1
2
𝐵 ∫
𝑑𝑣
𝑣
𝐼 = −𝐴 log 𝑢 +
1
2
𝐵 log 𝑣
𝐼 = −𝐴 log(3 − 𝑥) +
1
2
𝐵 log(3 + 2𝑥)
𝐼 = −
1
3
log(3 − 𝑥) +
1
2
 (−
1
3
) log(3 + 2𝑥)
𝐼 = −
1
3
log(3 − 𝑥) −
1
6
log(3 + 2𝑥)
9. (i) ∫
𝑥2 𝑑𝑥
(𝑥−𝑎)(𝑥−𝑏)(𝑥−𝑐)
(ii) ∫
𝑑𝑥
(𝑥−𝑎)2 (𝑥−𝑏)
(iii) ∫
𝑑𝑥
(𝑥−2)2(𝑥−1)3 (iv) ∫
𝑑𝑥
(𝑥+1)2(𝑥+2)3
(i) Solution: Let 𝐼 = ∫
𝑥2 𝑑𝑥
(𝑥−𝑎)(𝑥−𝑏)(𝑥−𝑐)
Let,
𝑥2
(𝑥−𝑎)(𝑥−𝑏)(𝑥−𝑐)
=
𝐴
(𝑥−𝑎)
+
𝐵
(𝑥−𝑏)
+
𝐶
(𝑥−𝑐)
------------- (a)
𝑥2
(𝑥−𝑎)(𝑥−𝑏)(𝑥−𝑐)
=
𝐴(𝑥−𝑏)(𝑥−𝑐)
(𝑥−𝑎)
+
𝐵(𝑥−𝑎)(𝑥−𝑐)
(𝑥−𝑏)
+
𝐶(𝑥−𝑎)(𝑥−𝑏)
(𝑥−𝑐)
𝑥2
(𝑥−𝑎)(𝑥−𝑏)(𝑥−𝑐)
=
𝐴(𝑥−𝑏)(𝑥−𝑐)+ 𝐵(𝑥−𝑎)(𝑥−𝑐)+𝐶(𝑥−𝑎)(𝑥−𝑏)
(𝑥−𝑎)(𝑥−𝑏)(𝑥−𝑐)

13. 𝑥2
= 𝐴(𝑥 − 𝑏)(𝑥 − 𝑐) + 𝐵(𝑥 − 𝑎)(𝑥 − 𝑐) + 𝐶(𝑥 − 𝑎)(𝑥 − 𝑏) --------- (b)
Putting, 𝑥 = 𝑎 , 𝑥 = 𝑏 and 𝑥 = 𝑐 in equation (b), we get
𝐴 =
𝑎2
(𝑎−𝑏)(𝑎−𝑐)
, 𝐵 =
𝑏2
(𝑏−𝑎)(𝑏−𝑐)
and 𝐶 =
𝑐2
(𝑐−𝑎)(𝑐−𝑏)
Now,
𝐼 = ∫
𝑥2 𝑑𝑥
(𝑥−𝑎)(𝑥−𝑏)(𝑥−𝑐)
𝐼 = ∫ {
𝐴
(𝑥−𝑎)
+
𝐵
(𝑥−𝑏)
+
𝐶
(𝑥−𝑐)
} 𝑑𝑥
𝐼 = ∫
𝐴
(𝑥−𝑎)
𝑑𝑥 + ∫
𝐵
(𝑥−𝑏)
𝑑𝑥 + ∫
𝐶
(𝑥−𝑐)
𝑑𝑥
𝐼 = 𝐴 ∫
𝑑𝑥
(𝑥−𝑎)
+ 𝐵 ∫
𝑑𝑥
(𝑥−𝑏)
+ 𝐶 ∫
𝑑𝑥
(𝑥−𝑐)
𝐼 = 𝐴 log(𝑥 − 𝑎) + 𝐵 log(𝑥 − 𝑏) + 𝐶 log(𝑥 − 𝑐)
𝐼 =
𝑎2
(𝑎−𝑏)(𝑎−𝑐)
log(𝑥 − 𝑎) +
𝑏2
(𝑏−𝑎)(𝑏−𝑐)
log(𝑥 − 𝑏) +
𝑐2
(𝑐−𝑎)(𝑐−𝑏)
log(𝑥 − 𝑐)
(ii) Solution: Let, 𝐼 = ∫
𝑑𝑥
(𝑥−𝑎)2 (𝑥−𝑏)
Let,
1
(𝑥−𝑎)2 (𝑥−𝑏)
=
𝐴
(𝑥−𝑎)
+
𝐵
(𝑥−𝑎)2 +
𝐶
(𝑥−𝑏)
--------- (a)
1
(𝑥−𝑎)2 (𝑥−𝑏)
=
𝐴 (𝑥−𝑎)(𝑥−𝑏)+ 𝐵(𝑥−𝑏)+ 𝐶 (𝑥−𝑎)2
(𝑥−𝑎)2 (𝑥−𝑏)
1 = 𝐴 (𝑥 − 𝑎)(𝑥 − 𝑏) + 𝐵(𝑥 − 𝑏) + 𝐶 (𝑥 − 𝑎)2
--------- (b)
Putting 𝑥 = 𝑎 and 𝑥 = 𝑏 in equation (b), we get
𝐵 =
1
𝑎−𝑏
and 𝐶 =
1
(𝑏−𝑎)2 -------- (c)
𝐴𝑔𝑎𝑖𝑛, From equation (b), we have
1 = 𝐴 (𝑥 − 𝑎)(𝑥 − 𝑏) + 𝐵(𝑥 − 𝑏) + 𝐶 (𝑥 − 𝑎)2
1 = 𝐴 (𝑥2
− (𝑎 + 𝑏)𝑥 + 𝑎𝑏) + 𝐵𝑥 − 𝑏𝐵 + 𝐶 (𝑥2
− 2𝑎𝑥 + 𝑎2)
1 = 𝐴 𝑥2
− (𝑎 + 𝑏)𝐴𝑥 + 𝑎𝑏𝐴 + 𝐵𝑥 − 𝑏𝐵 + 𝐶 𝑥2
− 2𝑎𝐶𝑥 + 𝑎2
𝐶 ---- (d)
Equating the constant terms on both sides of equation (d), we get
1 = 𝑎𝑏𝐴 − 𝑏𝐵 + 𝑎2
𝐶
1 = 𝑎𝑏𝐴 − 𝑏
1
𝑎−𝑏
+ 𝑎2

1
(𝑏−𝑎)2 Using (c)

14. 1 = 𝑎𝑏𝐴 −
𝑏
𝑎−𝑏
+
𝑎2
(𝑏−𝑎)2
𝑎𝑏𝐴 −
𝑏
𝑎−𝑏
+
𝑎2
(𝑏−𝑎)2 = 1
𝑎𝑏𝐴 +
𝑎2
(𝑏−𝑎)2 = 1 +
𝑏
𝑎−𝑏
𝑎𝑏𝐴 +
𝑎2
(𝑏−𝑎)2 =
𝑎+𝑏−𝑏
𝑎−𝑏
𝑎𝑏𝐴 +
𝑎2
(𝑏−𝑎)2 =
𝑎
𝑎−𝑏
𝑎𝑏𝐴 =
𝑎
𝑎−𝑏
−
𝑎2
(𝑏−𝑎)2
𝑎𝑏𝐴 =
𝑎
𝑎−𝑏
−
𝑎2
(𝑎−𝑏)2
𝑎𝑏𝐴 =
𝑎(𝑎−𝑏)−𝑎2
(𝑎−𝑏)2
𝑎𝑏𝐴 =
𝑎2−𝑎𝑏−𝑎2
(𝑎−𝑏)2
𝑎𝑏𝐴 =
−𝑎𝑏
(𝑎−𝑏)2
 𝐴 =
−1
(𝑎−𝑏)2
Now,
𝐼 = ∫
𝑑𝑥
(𝑥−𝑎)2 (𝑥−𝑏)
𝐼 = ∫ {
𝐴
(𝑥−𝑎)
+
𝐵
(𝑥−𝑎)2 +
𝐶
(𝑥−𝑏)
} 𝑑𝑥
𝐼 = ∫
𝐴
(𝑥−𝑎)
𝑑𝑥 + ∫
𝐵
(𝑥−𝑎)2 𝑑𝑥 + ∫
𝐶
(𝑥−𝑏)
𝑑𝑥
𝐼 = 𝐴 ∫
𝑑𝑥
(𝑥−𝑎)
+ 𝐵 ∫
1
(𝑥−𝑎)2 𝑑𝑥 + 𝐶 ∫
𝑑𝑥
(𝑥−𝑏)
𝐼 = 𝐴 ∫
𝑑𝑥
(𝑥−𝑎)
+ 𝐵 ∫(𝑥 − 𝑎)−2
𝑑𝑥 + 𝐶 ∫
𝑑𝑥
(𝑥−𝑏)
𝐼 = 𝐴 log(𝑥 − 𝑎) − 𝐵(𝑥 − 𝑎)−1
+ 𝐶 log(𝑥 − 𝑏)
𝐼 = −
1
(𝑎−𝑏)2 log(𝑥 − 𝑎) −
1
𝑎−𝑏
1
(𝑥−𝑎)
+
1
(𝑏−𝑎)2 log(𝑥 − 𝑏)
𝐼 =
1
(𝑎−𝑏)2
{log(𝑥 − 𝑏) − log(𝑥 − 𝑎)} −
1
𝑎−𝑏
1
(𝑥−𝑎)
𝐼 =
1
(𝑎−𝑏)2
{log
𝑥−𝑏
𝑥−𝑎
} −
1
𝑎−𝑏
1
(𝑥−𝑎)
𝐼 =
1
(𝑎−𝑏)2 log
𝑥−𝑏
𝑥−𝑎
−
1
−(𝑏−𝑎)
1
(𝑥−𝑎)

15. 𝐼 =
1
(𝑎−𝑏)2 log
𝑥−𝑏
𝑥−𝑎
+
1
(𝑏−𝑎)
1
(𝑥−𝑎)
𝐼 =
1
(𝑎−𝑏)2 log
𝑥−𝑏
𝑥−𝑎
+
1
(𝑏−𝑎)(𝑥−𝑎)
(iii) Solution: Let 𝐼 = ∫
𝑑𝑥
(𝑥−2)2(𝑥−1)3
{Integral of the form ∫
𝒅𝒙
(𝒙−𝒂)𝒎(𝒙−𝒃)𝒏 , can be evaluated by putting 𝒙 − 𝒂 = 𝒖(𝒙 − 𝒃)}.
Let, 𝑥 − 2 = 𝑢(𝑥 − 1) and differentiating with respect to 𝑥, we get
𝑥 − 2 = 𝑢𝑥 − 𝑢
𝑥 − 𝑢𝑥 = 2 − 𝑢
𝑥(1 − 𝑢) = 2 − 𝑢
 𝑥 =
2−𝑢
1−𝑢
-------- (a)
Differentiating with respect to 𝑢, we get
𝑑𝑥
𝑑𝑢
=
(1−𝑢)
𝑑
𝑑𝑢
(2−𝑢)−(2−𝑢)
𝑑
𝑑𝑢
(1−𝑢)
(1−𝑢)2
𝑑𝑥
𝑑𝑢
=
−(1−𝑢)+(2−𝑢)
(1−𝑢)2
𝑑𝑥
𝑑𝑢
=
−1+𝑢+2−𝑢
(1−𝑢)2
𝑑𝑥
𝑑𝑢
=
1
(1−𝑢)2
𝑑𝑥 =
𝑑𝑢
(1−𝑢)2
Now,
𝐼 = ∫
𝑑𝑥
(𝑥−2)2(𝑥−1)3
𝐼 = ∫
𝑑𝑢
(1−𝑢)2
(
2−𝑢
1−𝑢
−2)
2
(
2−𝑢
1−𝑢
−1)
3 Since, 𝑥 =
2−𝑢
1−𝑢
𝐼 = ∫
𝑑𝑢
(1−𝑢)2
(
2−𝑢−2(1−𝑢)
1−𝑢
−)
2
(
2−𝑢−(1−𝑢)
1−𝑢
)
3
𝐼 = ∫
𝑑𝑢
(1−𝑢)2
(
2−𝑢−2+2𝑢
1−𝑢
−)
2
(
2−𝑢−1+𝑢
1−𝑢
)
3

16. 𝐼 = ∫
𝑑𝑢
(1−𝑢)2
𝑢2
(1−𝑢)2
1
(1−𝑢)3
𝐼 = ∫
𝑑𝑢
(1−𝑢)2
𝑢2
(1−𝑢)5
𝐼 = ∫
𝑑𝑢
(1−𝑢)2 
(1−𝑢)5
𝑢2
𝐼 = ∫
(1−𝑢)3
𝑢2 𝑑𝑢
𝐼 = ∫
1−3𝑢+3𝑢2−𝑢3
𝑢2 𝑑𝑢
𝐼 = ∫
1
𝑢2 𝑑𝑢 − ∫
3𝑢
𝑢2 𝑑𝑢 + ∫
3𝑢2
𝑢2 𝑑𝑢 − ∫
𝑢3
𝑢2 𝑑𝑢
𝐼 = ∫ 𝑢−2
𝑑𝑢 − 3 ∫
1
𝑢
𝑑𝑢 + 3 ∫ 1 𝑑𝑢 − ∫ 𝑢 𝑑𝑢
𝐼 = −
1
𝑢
− 3 log 𝑢 + 3𝑢 −
1
2
𝑢2
We take, 𝑥 − 2 = 𝑢(𝑥 − 1)
 𝑢 =
𝑥−2
𝑥−1
Thus,
𝐼 = −
1
𝑢
− 3 log 𝑢 + 3𝑢 −
1
2
𝑢2
𝐼 = −
1
𝑥−2
𝑥−1
− 3 log
𝑥−2
𝑥−1
+ 3
𝑥−2
𝑥−1
−
1
2
(
𝑥−2
𝑥−1
)
2
𝐼 = −
𝑥−1
𝑥−2
− 3 log
𝑥−2
𝑥−1
+ 3
𝑥−2
𝑥−1
−
1
2
(
𝑥−2
𝑥−1
)
2
(iv) Solution: Let 𝐼 = ∫
𝑑𝑥
(𝑥+1)2(𝑥+2)3
Let, 𝑥 + 1 = 𝑢(𝑥 + 2)
𝑥 + 1 = 𝑢𝑥 + 2𝑢
𝑥 − 𝑢𝑥 = 2𝑢 − 1
𝑥(1 − 𝑢) = 2𝑢 − 1
 𝑥 =
2𝑢−1
1−𝑢

17. Differentiating with respect to 𝑢 , we get
𝑑𝑥
𝑑𝑢
=
(1−𝑢)
𝑑
𝑑𝑢
(2𝑢−1)−(2𝑢−1)
𝑑
𝑑𝑢
(1−𝑢)
(1−𝑢)2
𝑑𝑥
𝑑𝑢
=
2(1−𝑢)+(2𝑢−1)
(1−𝑢)2
𝑑𝑥
𝑑𝑢
=
2−2𝑢+2𝑢−1
(1−𝑢)2
𝑑𝑥
𝑑𝑢
=
1
(1−𝑢)2
 𝑑𝑥 =
𝑑𝑢
(1−𝑢)2
Now,
𝐼 = ∫
𝑑𝑥
(𝑥+1)2(𝑥+2)3
𝐼 = ∫
𝑑𝑢
(1−𝑢)2
(
2𝑢−1
1−𝑢
+1)
2
(
2𝑢−1
1−𝑢
+2)
3 Since, 𝑥 =
2𝑢−1
1−𝑢
𝐼 = ∫
𝑑𝑢
(1−𝑢)2
(
2𝑢−1+1−u
1−𝑢
)
2
(
2𝑢−1+2−2𝑢
1−𝑢
)
3
𝐼 = ∫
𝑑𝑢
(1−𝑢)2
(
u
1−𝑢
)
2
(
1
1−𝑢
)
3
𝐼 = ∫
𝑑𝑢
(1−𝑢)2
𝑢2
(1−𝑢)2
1
(1−𝑢)3
𝐼 = ∫
𝑑𝑢
(1−𝑢)2
𝑢2
(1−𝑢)5
𝐼 = ∫
𝑑𝑢
(1−𝑢)2 
(1−𝑢)5
𝑢2
𝐼 = ∫
(1−𝑢)3
𝑢2 𝑑𝑢
𝐼 = ∫
1−3𝑢+3𝑢2−𝑢3
𝑢2 𝑑𝑢
𝐼 = ∫
1
𝑢2 𝑑𝑢 − ∫
3𝑢
𝑢2 𝑑𝑢 + 3 ∫
𝑢2
𝑢2 𝑑𝑢 − ∫
𝑢3
𝑢2 𝑑𝑢
𝐼 = ∫ 𝑢−2
𝑑𝑢 − 3 ∫
1
𝑢
𝑑𝑢 + 3 ∫ 1 𝑑𝑢 − ∫ 𝑢 𝑑𝑢
𝐼 = −𝑢−1
− 3 log 𝑢 + 3𝑢 −
1
2
𝑢2
𝐼 = −
1
𝑢
− 3 log 𝑢 + 3𝑢 −
1
2
𝑢2

18. We take, 𝑥 + 1 = 𝑢(𝑥 + 2)
 𝑢 =
𝑥+1
𝑥+2
𝐼 = −
1
𝑥+1
𝑥+2
− 3 log
𝑥+1
𝑥+2
+ 3
𝑥+1
𝑥+2
−
1
2
(
𝑥+1
𝑥+2
)
2
𝐼 = −
𝑥+2
𝑥+1
− 3 log
𝑥+1
𝑥+2
+ 3
𝑥+1
𝑥+2
−
1
2
(
𝑥+1
𝑥+2
)
2
10. (i) ∫
𝑥2 𝑑𝑥
(𝑥+1)(𝑥+2)2 (ii) ∫
(3−𝑥)
𝑥2+𝑥3 𝑑𝑥
Solution: Let, 𝐼 = ∫
𝑥2 𝑑𝑥
(𝑥+1)(𝑥+2)2
Let,
𝑥2
(𝑥+1)(𝑥+2)2 =
𝐴
(𝑥+1)
+
𝐵
(𝑥+2)
+
𝐶
(𝑥+2)2 ------ (a)
𝑥2
(𝑥+1)(𝑥+2)2 =
𝐴(𝑥+2)2+𝐵(𝑥+1)(𝑥+2)+ 𝐶 (𝑥+1)
(𝑥+1)(𝑥+2)2
𝑥2
= 𝐴(𝑥 + 2)2
+ 𝐵(𝑥 + 1)(𝑥 + 2) + 𝐶 (𝑥 + 1) ----------- (b)
Putting, 𝑥 = −1 and 𝑥 = −2 in equation (b), we get
𝐴 = 1 , 𝐶 = −4
Again, from (b), we have
𝑥2
= 𝐴(𝑥2
+ 4𝑥 + 4) + 𝐵(𝑥2
+ 3𝑥 + 2) + 𝐶 (𝑥 + 1) ----- (c)
Equating the coefficient of 𝑥2
on both sides of equation (c), we get
1 = 𝐴 + 𝐵
1 = 1 + 𝐵 Since, 𝐴 = 1
 𝐵 = 0
Now,
𝐼 = ∫
𝑥2 𝑑𝑥
(𝑥+1)(𝑥+2)2
𝐼 = ∫ {
𝐴
(𝑥+1)
+
𝐵
(𝑥+2)
+
𝐶
(𝑥+2)2
} 𝑑𝑥 Using (a)
𝐼 = ∫
𝐴
(𝑥+1)
𝑑𝑥 + ∫
𝐵
(𝑥+2)
𝑑𝑥 + ∫
𝐶
(𝑥+2)2 𝑑𝑥
𝐼 = 𝐴 ∫
𝑑𝑥
(𝑥+1)
+ 𝐵 ∫
𝑑𝑥
(𝑥+2)
+ 𝐶 ∫
𝑑𝑥
(𝑥+2)2

19. 𝐼 = 𝐴 log(𝑥 + 1) + 𝐵 log(𝑥 + 2) + 𝐶 ∫(𝑥 + 2)−2
𝑑𝑥
𝐼 = 𝐴 log(𝑥 + 1) + 𝐵 log(𝑥 + 2) − 𝐶(𝑥 + 2)−1
𝐼 = log(𝑥 + 1) + 0 log(𝑥 + 2) − (−4)
1
(𝑥+2)
𝐼 = log(𝑥 + 1) +
4
(𝑥+2)
(iii) Solution: Let 𝐼 = ∫
(3−𝑥)
𝑥2+𝑥3 𝑑𝑥
𝐼 = ∫
(3−𝑥)
𝑥2(1+𝑥)
𝑑𝑥
Let,
(3−𝑥)
𝑥2(1+𝑥)
=
𝐴
𝑥
+
𝐵
𝑥2 +
𝐶
(1+𝑥)
------- (a)
(3−𝑥)
𝑥2(1+𝑥)
=
𝐴𝑥(1+𝑥)+ 𝐵(1+𝑥)+ 𝐶𝑥2
𝑥2(1+𝑥)
(3 − 𝑥) = 𝐴𝑥(1 + 𝑥) + 𝐵(1 + 𝑥) + 𝐶𝑥2
-------- (b)
Putting, 𝑥 = 0 and 𝑥 = −1 in equation (b), we get
𝐵 = 3 and 𝐶 = 4
Again, From (b), we have
(3 − 𝑥) = 𝐴𝑥(1 + 𝑥) + 𝐵(1 + 𝑥) + 𝐶𝑥2
(3 − 𝑥) = 𝐴(𝑥 + 𝑥2) + 𝐵(1 + 𝑥) + 𝐶𝑥2
Comparing the coefficients of 𝑥2
on both sides of equation (c), we get
0 = 𝐴 + 𝐶
𝐴 = −𝐶
 𝐴 = −4
Now,
𝐼 = ∫
(3−𝑥)
𝑥2(1+𝑥)
𝑑𝑥
𝐼 = ∫ {
𝐴
𝑥
+
𝐵
𝑥2 +
𝐶
(1+𝑥)
} 𝑑𝑥 Using (a)
𝐼 = ∫
𝐴
𝑥
𝑑𝑥 + ∫
𝐵
𝑥2 𝑑𝑥 + ∫
𝐶
(1+𝑥)
𝑑𝑥
𝐼 = 𝐴 ∫
𝑑𝑥
𝑥
+ 𝐵 ∫ 𝑥−2
𝑑𝑥 + 𝐶 ∫
𝑑𝑥
(1+𝑥)
𝐼 = 𝐴 log 𝑥 − 𝐵 𝑥−1
+ 𝐶 log(1 + 𝑥)

20. 𝐼 = 𝐴 log 𝑥 −
𝐵
𝑥
+ 𝐶 log(1 + 𝑥)
𝐼 = −4 log 𝑥 −
3
𝑥
+ 4 log(1 + 𝑥)
11. (i) ∫
𝑑𝑥
𝑥3−𝑥2−𝑥+1
(ii) ∫
𝑑𝑥
𝑥 (𝑥+1)2
(i) Solution: Let, 𝐼 = ∫
𝑑𝑥
𝑥3−𝑥2−𝑥+1
𝐼 = ∫
𝑑𝑥
𝑥2(𝑥−1)−𝑥+1
𝐼 = ∫
𝑑𝑥
𝑥2(𝑥−1)−1(𝑥−1)
𝐼 = ∫
𝑑𝑥
(𝑥−1)(𝑥2−1)
𝐼 = ∫
𝑑𝑥
(𝑥−1)(𝑥−1)(𝑥+1)
𝐼 = ∫
𝑑𝑥
(𝑥+1) (𝑥−1)2
Let,
1
(𝑥+1) (𝑥−1)2 =
𝐴
(𝑥+1)
+
𝐵
(𝑥−1)
+
𝐶
(𝑥−1)2 ------- (a)
1
(𝑥+1) (𝑥−1)2 =
𝐴(𝑥−1)2+ 𝐵(𝑥−1)(𝑥+1)+ 𝐶(𝑥+1)
(𝑥+1) (𝑥−1)2
1 = 𝐴(𝑥 − 1)2
+ 𝐵(𝑥 − 1)(𝑥 + 1) + 𝐶(𝑥 + 1) -------- (b)
Putting, 𝑥 = 1 and 𝑥 = −1 in equation (b), we get
𝐶 =
1
2
and 𝐴 =
1
4
Again, from (b), we have
1 = 𝐴(𝑥 − 1)2
+ 𝐵(𝑥 − 1)(𝑥 + 1) + 𝐶(𝑥 + 1)
1 = 𝐴 (𝑥2
− 2𝑥 + 1) + 𝐵(𝑥2
− 1) + 𝐶(𝑥 + 1) -------- (c)
Equating the constant terms on both sides of equation (c), we get
1 = 𝐴 − 𝐵 + 𝐶
1 =
1
4
− 𝐵 +
1
2
1 =
3
4
− 𝐵
𝐵 =
3
4
− 1
𝐵 = −
1
4

21. Now,
𝐼 = ∫
𝑑𝑥
(𝑥+1) (𝑥−1)2
𝐼 = ∫ {
𝐴
(𝑥+1)
+
𝐵
(𝑥−1)
+
𝐶
(𝑥−1)2
} 𝑑𝑥 Using (a)
𝐼 = ∫
𝐴
(𝑥+1)
𝑑𝑥 + ∫
𝐵
(𝑥−1)
𝑑𝑥 + ∫
𝐶
(𝑥−1)2 𝑑𝑥
𝐼 = 𝐴 ∫
𝑑𝑥
(𝑥+1)
+ 𝐵 ∫
𝑑𝑥
(𝑥−1)
+ 𝐶 ∫
1
(𝑥−1)2 𝑑𝑥
𝐼 = 𝐴 log(𝑥 + 1) + 𝐵 log(𝑥 − 1) + 𝐶 ∫(𝑥 − 1)−2
𝑑𝑥
𝐼 = 𝐴 log(𝑥 + 1) + 𝐵 log(𝑥 − 1) − 𝐶 (𝑥 − 1)−1
𝐼 = 𝐴 log(𝑥 + 1) + 𝐵 log(𝑥 − 1) −
𝐶
(𝑥−1)
𝐼 =
1
4
log(𝑥 + 1) −
1
4
log(𝑥 − 1) −
1
2
(𝑥−1)
𝐼 =
1
4
{log(𝑥 + 1) − log(𝑥 − 1)} −
1
2 (𝑥−1)
𝐼 =
1
4
log
𝑥+1
𝑥−1
−
1
2 (𝑥−1)
(ii) Solution: Let, 𝐼 = ∫
𝑑𝑥
𝑥 (𝑥+1)2
Let,
1
𝑥 (𝑥+1)2 =
𝐴
𝑥
+
𝐵
(𝑥+1)
+
𝐶
(𝑥+1)2 ----- (a)
1
𝑥 (𝑥+1)2 =
𝐴(𝑥+1)2+ 𝐵 𝑥 (𝑥+1)+ 𝐶𝑥
𝑥 (𝑥+1)2
1 = 𝐴(𝑥 + 1)2
+ 𝐵 𝑥 (𝑥 + 1) + 𝐶𝑥 ------- (b)
Putting, 𝑥 = 0 and 𝑥 = −1 in equation (b), we get
𝐴 = 1 and 𝐶 = −1
Again, from (b), we have
1 = 𝐴(𝑥 + 1)2
+ 𝐵 𝑥 (𝑥 + 1) + 𝐶𝑥
1 = 𝐴(𝑥2
+ 2𝑥 + 1) + 𝐵 (𝑥2
+ 𝑥) + 𝐶𝑥 ------ (c)
Equating the coefficients of 𝑥2
on both sides of equation (c), we get
0 = 𝐴 + 𝐵
𝐵 = −𝐴
𝐵 = −1

22. Now,
𝐼 = ∫
𝑑𝑥
𝑥 (𝑥+1)2
𝐼 = ∫ {
𝐴
𝑥
+
𝐵
(𝑥+1)
+
𝐶
(𝑥+1)2
} 𝑑𝑥
𝐼 = ∫
𝐴
𝑥
𝑑𝑥 + ∫
𝐵
(𝑥+1)
𝑑𝑥 + ∫
𝐶
(𝑥+1)2 𝑑𝑥
𝐼 = 𝐵 ∫
𝑑𝑥
𝑥
+ 𝐵 ∫
𝑑𝑥
(𝑥+1)
+ 𝐶 ∫
1
(𝑥+1)2 𝑑𝑥
𝐼 = 𝐴 log 𝑥 + 𝐵 log(𝑥 + 1) + 𝐶 ∫(𝑥 + 1)−2
𝑑𝑥
𝐼 = 𝐴 log 𝑥 + 𝐵 log(𝑥 + 1) − 𝐶(𝑥 + 1)−1
𝐼 = log 𝑥 − log(𝑥 + 1) −
−1
(𝑥+1)
𝐼 = log
𝑥
𝑥+1
+
1
(𝑥+1)
12. (i) ∫
𝑑𝑥
(𝑥2−1)2 (ii) ∫
(𝑥+1) 𝑑𝑥
(𝑥−1)2(𝑥+2)2
(i) Solution: Let 𝐼 = ∫
𝑑𝑥
(𝑥2−1)2
𝐼 = ∫
𝑑𝑥
(𝑥2−1)(𝑥2−1)
𝐼 = ∫
𝑑𝑥
(𝑥+1)(𝑥−1)(𝑥+1)(𝑥−1)
𝐼 = ∫
𝑑𝑥
(𝑥+1)2(𝑥−1)2
Let,
1
(𝑥+1)2(𝑥−1)2 =
𝐴
(𝑥+1)
+
𝐵
(𝑥+1)2 +
𝐶
(𝑥−1)
+
𝐷
(𝑥−1)2 -------- (a)
1
(𝑥+1)2(𝑥−1)2 =
𝐴(𝑥+1) (𝑥−1)2+ 𝐵(𝑥−1)2+ 𝐶(𝑥−1) (𝑥+1)2+ 𝐷 (𝑥+1)2
(𝑥+1)2(𝑥−1)2
1 = 𝐴(𝑥 + 1) (𝑥 − 1)2
+ 𝐵(𝑥 − 1)2
+ 𝐶(𝑥 − 1) (𝑥 + 1)2
+ 𝐷 (𝑥 + 1)2
--------- (b)
Putting, 𝑥 = 1 and 𝑥 = −1, we get
𝐷 =
1
4
and 𝐵 =
1
4
Again, from (b), we have
1 = 𝐴(𝑥 + 1) (𝑥 − 1)2
+ 𝐵(𝑥 − 1)2
+ 𝐶(𝑥 − 1) (𝑥 + 1)2
+ 𝐷 (𝑥 + 1)2
1 = 𝐴(𝑥 + 1) (𝑥2
− 2𝑥 + 1) + 𝐵(𝑥2
− 2𝑥 + 1) + 𝐶(𝑥 − 1) (𝑥2
+ 2𝑥 + 1) + 𝐷 (𝑥2
+ 2𝑥 + 1)

23. 1 = 𝐴 (𝑥3
− 𝑥2
− 𝑥 + 1) + 𝐵(𝑥2
− 2𝑥 + 1) + 𝐶 (𝑥3
+ 𝑥2
− 𝑥 − 1) + 𝐷 (𝑥2
+ 2𝑥 + 1) ---- (c)
Equating the constant terms on both sides of equation (c), we get
1 = 𝐴 + 𝐵 − 𝐶 + 𝐷
1 = 𝐴 +
1
4
− 𝐶 +
1
4
1 = 𝐴 − 𝐶 +
1
2
𝐴 − 𝐶 =
1
2
𝐴 = 𝐶 +
1
2
------- (d)
Again, equating the coefficients of 𝑥2
on both sides of equation (c), we get
0 = −𝐴— 2𝐵 − 𝐶 + 2𝐷
 𝐶 = −𝐴 − 2𝐵 + 2𝐷 --------- (e)
Solving (d) and (e), we get
𝐶 = −
1
4
Again, putting 𝐶 = −
1
4
in (d), we get
𝐴 =
1
4
Now,
𝐼 = ∫
𝑑𝑥
(𝑥+1)2(𝑥−1)2
𝐼 = ∫ {
𝐴
(𝑥+1)
+
𝐵
(𝑥+1)2 +
𝐶
(𝑥−1)
+
𝐷
(𝑥−1)2
} 𝑑𝑥 Using (a)
𝐼 = ∫
𝐴
(𝑥+1)
𝑑𝑥 + ∫
𝐵
(𝑥+1)2 𝑑𝑥 + ∫
𝐶
(𝑥−1)
𝑑𝑥 + ∫
𝐷
(𝑥−1)2 𝑑𝑥
𝐼 = 𝐴 ∫
𝑑𝑥
(𝑥+1)
+ 𝐵 ∫(𝑥 + 1)−2
𝑑𝑥 + 𝐶 ∫
𝑑𝑥
(𝑥−1)
+ 𝐷 ∫(𝑥 − 1)−2
𝑑𝑥
𝐼 = 𝐴 log(𝑥 + 1) − 𝐵(𝑥 + 1)−1
+ 𝐶 log(𝑥 − 1 ) − 𝐷(𝑥 − 1)−1
𝐼 = 𝐴 log(𝑥 + 1) −
𝐵
(𝑥+1)
+ 𝐶 log(𝑥 − 1 ) −
𝐷
(𝑥−1)
𝐼 =
1
4
log(𝑥 + 1) −
1
4
(𝑥+1)
−
1
4
log(𝑥 − 1 ) −
1
4
(𝑥−1)
𝐼 =
1
4
{log(𝑥 + 1) − log(𝑥 − 1)} −
1
4(𝑥+1)
−
1
4(𝑥−1)
𝐼 =
1
4
log
𝑥+1
𝑥−1
−
1
4
(
(𝑥−1)+(𝑥+1)
(𝑥+1)(𝑥−1)
)

24. 𝐼 =
1
4
log
𝑥+1
𝑥−1
−
1
4
(
2𝑥
𝑥2−1
)
𝐼 =
1
4
log
𝑥+1
𝑥−1
−
1
2
(
𝑥
𝑥2−1
)
(ii) Solution: Let 𝐼 = ∫
(𝑥+1) 𝑑𝑥
(𝑥−1)2(𝑥+2)2
Let,
(𝑥+1)
(𝑥−1)2(𝑥+2)2 =
𝐴
(𝑥−1)
+
𝐵
(𝑥−1)2 +
𝐶
(𝑥+2)
+
𝐷
(𝑥+2)2 ------- (b)
(𝑥+1)
(𝑥−1)2(𝑥+2)2 =
𝐴(𝑥−1)(𝑥+2)2+ 𝐵(𝑥+2)2+ 𝐶 (𝑥−1)2(𝑥+2)+ 𝐷 (𝑥−1)2
(𝑥−1)2(𝑥+2)2
(𝑥 + 1) = 𝐴(𝑥 − 1)(𝑥 + 2)2
+ 𝐵(𝑥 + 2)2
+ 𝐶 (𝑥 − 1)2(𝑥 + 2) + 𝐷 (𝑥 − 1)2
----- (c)
Putting, 𝑥 = 1 and 𝑥 = −2 in equation (c), we get
𝐵 =
2
9
and 𝐷 = −
1
9
Again, from (c), we have
(𝑥 + 1) = 𝐴(𝑥 − 1)(𝑥 + 2)2
+ 𝐵(𝑥 + 2)2
+ 𝐶 (𝑥 − 1)2(𝑥 + 2) + 𝐷 (𝑥 − 1)2
𝑥 + 1 = 𝐴(𝑥 − 1)(𝑥2
+ 4𝑥 + 4) + 𝐵(𝑥2
+ 4𝑥 + 4) + 𝐶 (𝑥2
− 2𝑥 + 1)(𝑥 + 2) + 𝐷 (𝑥2
− 2𝑥 + 1)
𝑥 + 1 = 𝐴(𝑥3
+ 3𝑥2
− 4) + 𝐵(𝑥2
+ 4𝑥 + 4) + 𝐶 (𝑥3
− 3𝑥 + 2) + 𝐷 (𝑥2
− 2𝑥 + 1) ----- (d)
Equating the coefficients of 𝑥 on both sides of equation (d), we get
1 = 4𝐵 − 3𝐶 − 2𝐷
1 = 4
2
9
− 3𝐶 − 2 (−
1
9
)
1 =
8
9
+
2
9
− 3𝐶
1 =
10
9
− 3𝐶
3𝐶 =
10
9
− 1
3𝐶 =
1
9
𝐶 =
1
27
Again, equating the coefficients of 𝑥3
on both sides of (d), we get
0 = 𝐴 + 𝐶
𝐴 = −𝐶

25. 𝐴 = −
1
27
𝐻𝑒𝑛𝑐𝑒, 𝐴 = −
1
27
, , =
2
9
, 𝐶 =
1
27
and 𝐷 = −
1
9
Now,
𝐼 = ∫
(𝑥+1) 𝑑𝑥
(𝑥−1)2(𝑥+2)2
𝐼 = ∫ {
𝐴
(𝑥−1)
+
𝐵
(𝑥−1)2 +
𝐶
(𝑥+2)
+
𝐷
(𝑥+2)2
} 𝑑𝑥
𝐼 = ∫
𝐴
(𝑥−1)
𝑑𝑥 + ∫
𝐵
(𝑥−1)2 𝑑𝑥 + ∫
𝐶
(𝑥+2)
𝑑𝑥 + ∫
𝐷
(𝑥+2)2 𝑑𝑥
𝐼 = 𝐴 log(𝑥 − 1) + 𝐵 ∫(𝑥 − 1)−2
𝑑𝑥 + 𝐶 log(𝑥 + 2) + 𝐷 ∫(𝑥 + 2)−2
𝑑𝑥
𝐼 = 𝐴 log(𝑥 − 1) − 𝐵(𝑥 − 1)−1
+ 𝐶 log(𝑥 + 2) − 𝐷 (𝑥 + 2)−1
𝐼 = 𝐴 log(𝑥 − 1) −
𝐵
(𝑥−1)
+ 𝐶 log(𝑥 + 2) −
𝐷
(𝑥+2)
𝐼 = −
1
27
log(𝑥 − 1) −
2
9
(𝑥−1)
+
1
27
log(𝑥 + 2) −
−
1
9
(𝑥+2)
𝐼 = −
1
27
{log(𝑥 − 1) − log(𝑥 + 2)} +
1
9
{
1
(𝑥+2)
−
2
(𝑥−1)
}
𝐼 = −
1
27
log
𝑥−1
𝑥+2
+
1
9
{
1
(𝑥+2)
−
2
(𝑥−1)
}
𝐼 =
1
9
{
1
(𝑥+2)
−
2
(𝑥−1)
} −
1
27
log
𝑥−1
𝑥+2
𝐼 =
1
9
{
1
(𝑥+2)
−
2
(𝑥−1)
−
1
3
log
𝑥−1
𝑥+2
}
13. (i) ∫
𝑑𝑥
(𝑥−1)3 (𝑥+1)
(ii) ∫
(3𝑥+2) 𝑑𝑥
𝑥 (𝑥+1)3
(i) Solution: Let 𝐼 = ∫
𝑑𝑥
(𝑥−1)3 (𝑥+1)
Let,
1
(𝑥−1)3 (𝑥+1)
=
𝐴
(𝑥−1)
+
𝐵
(𝑥−1)2 +
𝐶
(𝑥−1)3 +
𝐷
(𝑥+1)
-------- (a)
1
(𝑥−1)3 (𝑥+1)
=
𝐴 (𝑥−1)2 (𝑥+1)+ 𝐵 (𝑥−1) (𝑥+1)+ 𝐶(𝑥+1)+ 𝐷 (𝑥−1)3
(𝑥−1)3 (𝑥+1)
1 = 𝐴 (𝑥 − 1)2 (𝑥 + 1) + 𝐵 (𝑥 − 1) (𝑥 + 1) + 𝐶(𝑥 + 1) + 𝐷 (𝑥 − 1)3
----- (b)
Putting, 𝑥 = 1 and 𝑥 = −1 in equation (b), we get

26. 𝐶 =
1
2
and 𝐷 = −
1
8
Again, from equation (b), we have
1 = 𝐴 (𝑥 − 1)2 (𝑥 + 1) + 𝐵 (𝑥 − 1) (𝑥 + 1) + 𝐶(𝑥 + 1) + 𝐷 (𝑥 − 1)3
1 = 𝐴 (𝑥2
− 2𝑥 + 1) (𝑥 + 1) + 𝐵 (𝑥2
− 1) + 𝐶(𝑥 + 1) + 𝐷 (𝑥3
− 3𝑥2
+ 3𝑥 − 1)
1 = 𝐴 (𝑥3
− 𝑥2
− 𝑥 + 1) + 𝐵 (𝑥2
− 1) + 𝐶(𝑥 + 1) + 𝐷 (𝑥3
− 3𝑥2
+ 3𝑥 − 1) ------ (c)
Equating the coefficients of 𝑥3
on both sides of equation (c), we get
0 = 𝐴 + 𝐷
𝐴 = −𝐷
𝐴 = − (−
1
8
)
𝐴 =
1
8
Also, equating the coefficients of 𝑥2
on both sides of equation (c), we get
0 = −𝐴 + 𝐵 − 3𝐷
𝐵 = 𝐴 + 3𝐷
𝐵 =
1
8
+ 3 (−
1
8
)
𝐵 =
1
8
−
3
8
𝐵 = −
2
8
𝐵 = −
1
4
𝐻𝑒𝑛𝑐𝑒, 𝐴 =
1
8
𝐵 = −
1
4
, 𝐶 =
1
2
and 𝐷 = −
1
8
Now,
𝐼 = ∫
𝑑𝑥
(𝑥−1)3 (𝑥+1)
𝐼 = ∫ {
𝐴
(𝑥−1)
+
𝐵
(𝑥−1)2 +
𝐶
(𝑥−1)3 +
𝐷
(𝑥+1)
} 𝑑𝑥 Using (a)
𝐼 = ∫
𝐴
(𝑥−1)
𝑑𝑥 + ∫
𝐵
(𝑥−1)2 𝑑𝑥 + ∫
𝐶
(𝑥−1)3 𝑑𝑥 + ∫
𝐷
(𝑥+1)
𝑑𝑥
𝐼 = 𝐴 ∫
𝑑𝑥
(𝑥−1)
+ 𝐵 ∫
1
(𝑥−1)2 𝑑𝑥 + 𝐶 ∫
1
(𝑥−1)3 𝑑𝑥 + 𝐷 ∫
𝑑𝑥
(𝑥+1)
𝐼 = 𝐴 log(𝑥 − 1) + 𝐵 ∫(𝑥 − 1)−2
𝑑𝑥 + 𝐶 ∫(𝑥 − 1)−3
𝑑𝑥 + 𝐷 log(𝑥 + 1)

27. 𝐼 = 𝐴 log(𝑥 − 1) + 𝐷 log(𝑥 + 1) − 𝐵(𝑥 − 1)−1
−
𝐶
2
(𝑥 − 1)−2
𝐼 =
1
8
log(𝑥 − 1) −
1
8
log(𝑥 + 1) +
1
4
1
(𝑥−1)
−
1
2
2

1
(𝑥−1)2
𝐼 =
1
8
{log(𝑥 − 1) − log(𝑥 + 1)} +
1
4
{
1
(𝑥−1)
−
1
(𝑥−1)2
}
𝐼 =
1
8
log
𝑥−1
𝑥+1
+
1
4
{
𝑥−1−1
(𝑥−1)2
}
𝐼 =
1
8
log
𝑥−1
𝑥+1
+
1
4
{
𝑥−2
(𝑥−1)2
}
(ii) Solution: Let, 𝐼 = ∫
(3𝑥+2) 𝑑𝑥
𝑥 (𝑥+1)3
Let,
(3𝑥+2)
𝑥 (𝑥+1)3 =
𝐴
𝑥
+
𝐵
(𝑥+1)
+
𝐶
(𝑥+1)2 +
𝐷
(𝑥+1)3 -------- (a)
(3𝑥+2)
𝑥 (𝑥+1)3 =
𝐴(𝑥+1)3 + 𝐵𝑥(𝑥+1)2 +𝐶 𝑥(𝑥+1)+ 𝐷𝑥
𝑥 (𝑥+1)3
(3𝑥 + 2) = 𝐴(𝑥 + 1)3
+ 𝐵𝑥(𝑥 + 1)2
+ 𝐶 𝑥(𝑥 + 1) + 𝐷𝑥 ------ (b)
Putting, = 0 , 𝑥 = −1 in equation (b), we get
𝐴 = 2 and 𝐷 = 1
Again, from equation (b), we have
(3𝑥 + 2) = 𝐴(𝑥 + 1)3
+ 𝐵𝑥(𝑥 + 1)2
+ 𝐶 𝑥(𝑥 + 1) + 𝐷𝑥
(3𝑥 + 2) = 𝐴(𝑥3
+ 3𝑥2
+ 3𝑥 + 1) + 𝐵(𝑥3
+ 2𝑥2
+ 𝑥) + 𝐶 (𝑥2
+ 𝑥) + 𝐷𝑥 ------ (c)
Equating the coefficients of 𝑥3
on both sides the equation (c), we get
0 = 𝐴 + 𝐵
𝐵 = −𝐴
𝐵 = −2
Equating the coefficients of 𝑥2
on both sides the equation (c), we get
0 = 3𝐴 + 2𝐵 + 𝐶
𝐶 = −3𝐴 − 2𝐵
𝐶 = −32 − 2(−2)
𝐶 = −6 + 4
𝐶 = −2

28. Hence, 𝐴 = 2 , = −2 , 𝐶 = −2 and 𝐷 = 1
Now,
𝐼 = ∫
(3𝑥+2) 𝑑𝑥
𝑥 (𝑥+1)3
𝐼 = ∫ {
𝐴
𝑥
+
𝐵
(𝑥+1)
+
𝐶
(𝑥+1)2 +
𝐷
(𝑥+1)3
} 𝑑𝑥
𝐼 = ∫
𝐴
𝑥
𝑑𝑥 + ∫
𝐵
(𝑥+1)
𝑑𝑥 + ∫
𝐶
(𝑥+1)2 𝑑𝑥 + ∫
𝐷
(𝑥+1)3 𝑑𝑥
𝐼 = 𝐴 ∫
𝑑𝑥
𝑥
+ 𝐵 ∫
𝑑𝑥
(𝑥+1)
+ 𝐶 ∫
1
(𝑥+1)2 𝑑𝑥 + 𝐷 ∫
1
(𝑥+1)3 𝑑𝑥
𝐼 = 𝐴 ∫
𝑑𝑥
𝑥
+ 𝐵 ∫
𝑑𝑥
(𝑥+1)
+ 𝐶 ∫(𝑥 + 1)−2
𝑑𝑥 + 𝐷 ∫(𝑥 + 1)−3
𝑑𝑥
𝐼 = 𝐴 log 𝑥 + 𝐵 log(𝑥 + 1) − 𝐶 (𝑥 + 1)−1
−
1
2
𝐷(𝑥 + 1)−2
𝐼 = 𝐴 log 𝑥 + 𝐵 log(𝑥 + 1) − 𝐶
1
(𝑥+1)
−
1
2
𝐷
1
(𝑥+1)2
𝐼 = 2 log 𝑥 − 2 log(𝑥 + 1) + 2
1
(𝑥+1)
−
1
2
1
(𝑥+1)2
𝐼 = 2{log 𝑥 − log(𝑥 + 1)} +
2
(𝑥+1)
−
1
2(𝑥+1)2
𝐼 = 2 log
𝑥
𝑥+1
+
4(𝑥+1)−1
2(𝑥+1)2
𝐼 = 2 log
𝑥
𝑥+1
+
4𝑥+3
2(𝑥+1)2
14. (i) ∫
𝑑𝑥
1−𝑥3 (ii) ∫
2+ 𝑥2
1−𝑥3 𝑑𝑥
(i) Solution: Let, 𝐼 = ∫
𝑑𝑥
1−𝑥3
Here, 1 − 𝑥3
= (1 − 𝑥) (1 + 𝑥 + 𝑥2)
Let,
1
1−𝑥3 =
𝐴
1−𝑥
+
𝐵𝑥+𝐶
1+𝑥+𝑥2 ------- (a)
1 = 𝐴(1 + 𝑥 + 𝑥2) + (𝐵𝑥 + 𝐶)(1 − 𝑥) -------- (b)
Putting, 𝑥 = 0 and 𝑥 = 1 in (b), we get
For, 𝑥 = 1 , 𝐴 =
1
3

29. For, 𝑥 = 0
𝐴 + 𝐶 = 1
𝐶 = 1 − 𝐴
 𝐶 = 1 −
1
3
=
2
3
Again, equating the coefficients of 𝑥2
, we get
0 = 𝐴 − 𝐵
𝐵 = 𝐴
 𝐵 =
1
3
Now,
𝐼 = ∫
𝑑𝑥
1−𝑥3
𝐼 = ∫ (
𝐴
1−𝑥
+
𝐵𝑥+𝐶
1+𝑥+𝑥2
) 𝑑𝑥 Using (a)
𝐼 = ∫
𝐴
1−𝑥
𝑑𝑥 + ∫
𝐵𝑥+𝐶
1+𝑥+𝑥2 𝑑𝑥
𝐼 = ∫
1
3
1−𝑥
𝑑𝑥 + ∫
1
3
𝑥+
2
3
1+𝑥+𝑥2 𝑑𝑥
𝐼 =
1
3
∫
𝑑𝑥
1−𝑥
+
1
3
∫
𝑥+2
1+𝑥+𝑥2 𝑑𝑥
𝐼 =
1
3
∫
𝑑𝑥
1−𝑥
+
1
3

1
2
∫
2𝑥+4
1+𝑥+𝑥2 𝑑𝑥
𝐼 =
1
3
∫
𝑑𝑥
1−𝑥
+
1
6
∫
2𝑥+1+3
1+𝑥+𝑥2 𝑑𝑥
𝐼 =
1
3
∫
𝑑𝑥
1−𝑥
+
1
6
∫
2𝑥+1
1+𝑥+𝑥2 𝑑𝑥 +
1
6
∫
3
1+𝑥+𝑥2 𝑑𝑥
𝐼 =
1
3
∫
𝑑𝑥
1−𝑥
+
1
6
∫
2𝑥+1
1+𝑥+𝑥2 𝑑𝑥 +
1
2
∫
𝑑𝑥
1+𝑥+𝑥2
𝐼 =
1
3
∫
−𝑑𝑢
𝑢
+
1
6
∫
𝑑𝑣
𝑣
𝑑𝑥 +
1
2
∫
𝑑𝑥
1+𝑥+𝑥2
𝐼 = −
1
3
log 𝑣 +
1
6
log 𝑣 +
1
2
∫
𝑑𝑥
1+𝑥+𝑥2
Putting,
1 − 𝑥 = 𝑢
 𝑑𝑥 = −𝑑𝑢
Putting,
1 + 𝑥 + 𝑥2
= 𝑣
 (2𝑥 + 1) 𝑑𝑥 = 𝑑𝑣

30. 𝐼 = −
1
3
log(1 − 𝑥) +
1
6
log(1 + 𝑥 + 𝑥2) +
1
2
∫
𝑑𝑥
1+𝑥+𝑥2
𝐼 = −
1
3
log(1 − 𝑥) +
1
3

1
2
log(1 + 𝑥 + 𝑥2) +
1
2
∫
𝑑𝑥
1+𝑥+𝑥2
𝐼 = −
1
3
log(1 − 𝑥) +
1
3
log(1 + 𝑥 + 𝑥2)
1
2 +
1
2
∫
𝑑𝑥
1+𝑥+𝑥2
𝐼 = −
1
3
log(1 − 𝑥) +
1
3
log(√1 + 𝑥 + 𝑥2) +
1
2
∫
𝑑𝑥
1+𝑥+𝑥2
𝐼 = −
1
3
{log(1 − 𝑥) − log √1 + 𝑥 + 𝑥2} +
1
2
∫
𝑑𝑥
1+𝑥+𝑥2
𝐼 = −
1
3
log
1−𝑥
√1+𝑥+𝑥2
+
1
2
∫
𝑑𝑥
1+𝑥+𝑥2
𝐼 = −
1
3
log
1−𝑥
√1+𝑥+𝑥2
+
1
2
∫
𝑑𝑥
(𝑥+
1
2
)
2
+(
√3
2
)
2
𝐼 = −
1
3
log
1−𝑥
√1+𝑥+𝑥2
+
1
2

1
√3
2
tan−1
(
𝑥+
1
2
√3
2
)
𝐼 = −
1
3
log
1−𝑥
√1+𝑥+𝑥2
+
1
√3
tan−1 (
2𝑥+1
√3
)
𝐼 =
1
√3
tan−1 (
2𝑥+1
√3
) −
1
3
log
1−𝑥
√1+𝑥+𝑥2
(ii) Solution: Let, 𝐼 = ∫
2+ 𝑥2
1−𝑥3 𝑑𝑥
Here, 1 − 𝑥3
= (1 − 𝑥)(1 + 𝑥 + 𝑥2)
Let,
2+ 𝑥2
1−𝑥3 =
𝐴
1−𝑥
+
𝐵𝑥+𝐶
1+𝑥+𝑥2 ------- (a)
2 + 𝑥2
= 𝐴(1 + 𝑥 + 𝑥2) + (𝐵𝑥 + 𝐶)(1 − 𝑥)
2 + 𝑥2
= 𝐴(1 + 𝑥 + 𝑥2) + 𝐵(𝑥 − 𝑥2) + 𝐶(1 − 𝑥) ------ (b)
Putting, 𝑥 = 1 in (b), we get
3 = 3𝐴  𝐴 = 1
Putting, 𝑥 = 0 in (b), we get
2 = 𝐴 + 𝐶
𝐶 = 2 − 𝐴
𝐶 = 2 − 1  𝐶 = 1
Equating the coefficients of 𝑥2
on both sides of equation (b), we get
1 = 𝐴 − 𝐵

31. 𝐵 = 𝐴 − 1
 𝐵 = 1 − 1 = 0
Now,
𝐼 = ∫
2+ 𝑥2
1−𝑥3 𝑑𝑥
𝐼 = ∫ (
𝐴
1−𝑥
+
𝐵𝑥+𝐶
1+𝑥+𝑥2
) 𝑑𝑥
𝐼 = ∫
𝐴
1−𝑥
𝑑𝑥 + ∫
𝐵𝑥+𝐶
1+𝑥+𝑥2 𝑑𝑥
𝐼 = 𝐴 ∫
𝑑𝑥
1−𝑥
+ 𝐶 ∫
𝑑𝑥
1+𝑥+𝑥2 Since, 𝐵 = 0
𝐼 = ∫
𝑑𝑥
1−𝑥
+ ∫
𝑑𝑥
(𝑥+
1
2
)
2
+(
√3
2
)
2
𝐼 = −log(1 − 𝑥) +
1
√3
2
tan−1
(
𝑥+
1
2
√3
2
)
𝐼 = −log(1 − 𝑥) +
2
√3
tan−1 (
2𝑥+1
√3
)
15. (i) ∫
𝑥
𝑥4−1
𝑑𝑥 (ii) ∫
𝑑𝑥
𝑥(𝑥4−1)
(i) Solution: Let 𝐼 = ∫
𝑥
𝑥4−1
𝑑𝑥
𝐼 = ∫
𝑥
(𝑥2−1)(𝑥2+1)
𝑑𝑥
Take, 𝑥2
= 𝑢  𝑥 𝑑𝑥 =
1
2
𝑑𝑢
Now,
𝐼 = ∫
𝑥
(𝑥2−1)(𝑥2+1)
𝑑𝑥
𝐼 = ∫
1
2
𝑑𝑢
(𝑢−1)(𝑢+1)
𝐼 =
1
2
∫
𝑑𝑢
(𝑢−1)(𝑢+1)
Let,
1
(𝑢−1)(𝑢+1)
=
𝐴
𝑢−1
+
𝐵
𝑢+1
-------- (a)
1 = 𝐴(𝑢 + 1) + 𝐵 (𝑢 − 1) ------ (b)
Putting, 𝑢 = 1 in (b), we get
1 = 2𝐴  𝐴 =
1
2

32. Putting, 𝑢 = −1 in (b), we get
1 = −2𝐵  𝐵 = −
1
2
Now,
𝐼 =
1
2
∫
𝑑𝑢
(𝑢−1)(𝑢+1)
𝐼 =
1
2
∫ (
𝐴
𝑢−1
+
𝐵
𝑢+1
) 𝑑𝑢 Using (a)
𝐼 =
1
2
∫
𝐴
𝑢−1
𝑑𝑢 +
1
2
∫
𝐵
𝑢+1
𝑑𝑢
𝐼 =
1
2
𝐴 ∫
𝑑𝑢
𝑢−1
+
1
2
𝐵 ∫
𝑑𝑢
𝑢+1
𝐼 =
1
2

1
2
log(𝑢 − 1) +
1
2
 (−
1
2
) log(𝑢 + 1)
𝐼 =
1
4
log(𝑢 − 1) −
1
4
log(𝑢 + 1)
𝐼 =
1
4
log(𝑥2
− 1) −
1
4
log(𝑥2
+ 1)
𝐼 =
1
4
{log(𝑥2
− 1) − log(𝑥2
+ 1)}
(ii) Solution: Let, 𝐼 = ∫
𝑑𝑥
𝑥(𝑥4−1)
𝐼 = ∫
𝑥3𝑑𝑥
𝑥3 𝑥(𝑥4−1)
𝐼 = ∫
𝑥3𝑑𝑥
𝑥4 (𝑥4−1)
Take, 𝑥4
= 𝑢  𝑥3
𝑑𝑥 =
1
4
𝑑𝑢
𝐼 = ∫
1
4
𝑑𝑢
𝑢 (𝑢−1)
𝐼 =
1
4
∫
𝑑𝑢
𝑢 (𝑢−1)
Let,
1
𝑢 (𝑢−1)
=
𝐴
𝑢
+
𝐵
𝑢−1
-------- (a)
1 = 𝐴(𝑢 − 1) + 𝐵𝑢 ------ (b)
Putting, 𝑢 = 𝑜 in (b), we get
𝐴 = −1
Putting, 𝑢 = 1 in (b), we get

33. 𝐵 = 1
Now,
𝐼 =
1
4
∫
𝑑𝑢
𝑢 (𝑢−1)
𝐼 =
1
4
∫ (
𝐴
𝑢
+
𝐵
𝑢−1
) 𝑑𝑢
𝐼 =
1
4
∫
𝐴
𝑢
𝑑𝑢 +
1
4
∫
𝐵
𝑢−1
𝑑𝑢
𝐼 =
1
4
𝐴 ∫
𝑑𝑢
𝑢
+
1
4
𝐵 ∫
𝑑𝑢
𝑢−1
𝐼 =
1
4
𝐴 log 𝑢 +
1
4
𝐵 log(𝑢 − 1)
𝐼 = −
1
4
log 𝑢 +
1
4
log(𝑢 − 1)
𝐼 = −
1
4
log 𝑥4
+
1
4
log(𝑥4
− 1)
𝐼 = −
4
4
log 𝑥 +
1
4
log(𝑥4
− 1)
𝐼 = −log 𝑥 +
1
4
log(𝑥4
− 1)
𝐼 =
1
4
log(𝑥4
− 1) − log 𝑥
16. (i) ∫
𝑥2
1−𝑥4 𝑑𝑥 (ii) ∫
𝑑𝑥
𝑥4−1
(i) Solution: Let, 𝐼 = ∫
𝑥2
1−𝑥4 𝑑𝑥
𝐼 =
1
2
∫
2𝑥2
1−𝑥4 𝑑𝑥
𝐼 =
1
2
∫
(1+𝑥2)−(1−𝑥2)
(1+𝑥2)(1−𝑥2)
𝑑𝑥
𝐼 =
1
2
∫ (
(1+𝑥2)
(1+𝑥2)(1−𝑥2)
−
(1−𝑥2)
(1+𝑥2)(1−𝑥2)
) 𝑑𝑥
𝐼 =
1
2
∫ (
1
(1−𝑥2)
−
1
(1+𝑥2)
) 𝑑𝑥
𝐼 =
1
2
∫
1
1−𝑥2 𝑑𝑥 −
1
2
∫
1
1+𝑥2 𝑑𝑥
𝐼 =
1
2
∫
1
(1+𝑥)(1−𝑥)
𝑑𝑥 −
1
2
tan−1
𝑥
Here,
1
(1+𝑥)(1−𝑥)
=
𝐴
(1+𝑥)
+
𝐵
(1−𝑥)
-------- (a)

34. 1
(1+𝑥)(1−𝑥)
=
𝐴(1−𝑥)+ 𝐵 (1+𝑥)
(1+𝑥)
1 = 𝐴(1 − 𝑥) + 𝐵 (1 + 𝑥) ------- (b)
Putting, 𝑥 = 1 in (b), we get
𝐵 =
1
2
Putting, 𝑥 = −1 in (b), we get
𝐴 =
1
2
𝑁𝑜𝑤,
𝐼 =
1
2
∫
1
(1+𝑥)(1−𝑥)
𝑑𝑥 −
1
2
tan−1
𝑥
𝐼 =
1
2
{∫ (
𝐴
(1+𝑥)
+
𝐵
(1−𝑥)
)𝑑𝑥} −
1
2
tan−1
𝑥 Using (a)
𝐼 =
1
2
{∫
𝐴
(1+𝑥)
𝑑𝑥 + ∫
𝐵
(1−𝑥)
𝑑𝑥} −
1
2
tan−1
𝑥
𝐼 =
1
2
{𝐴 ∫
𝑑𝑥
(1+𝑥)
+ 𝐵 ∫
𝑑𝑥
(1−𝑥)
} −
1
2
tan−1
𝑥
𝐼 =
1
2
{𝐴 log(1 + 𝑥) − 𝐵 log(1 − 𝑥)} −
1
2
tan−1
𝑥
𝐼 =
1
2
{
1
2
log(1 + 𝑥) −
1
2
log(1 − 𝑥)} −
1
2
tan−1
𝑥
𝐼 =
1
4
log(1 + 𝑥) −
1
4
log(1 − 𝑥) −
1
2
tan−1
𝑥
𝐼 =
1
4
{log(1 + 𝑥) − log(1 − 𝑥)} −
1
2
tan−1
𝑥
(ii) Solution: Let , 𝐼 = ∫
𝑑𝑥
𝑥4−1
𝐼 = ∫
𝑑𝑥
(𝑥+1)(𝑥−1)(𝑥2+1)
Here,
1
(𝑥+1)(𝑥−1)(𝑥2+1)
=
𝐴
(𝑥+1)
+
𝐵
(𝑥−1)
+
𝐶𝑥+𝐷
(𝑥2+1)
------- (a)
1
(𝑥+1)(𝑥−1)(𝑥2+1)
=
𝐴(𝑥−1)(𝑥2+1)+ 𝐵(𝑥+1)(𝑥2+1)+ (𝐶𝑥+𝐷) (𝑥+1)(𝑥−1)
(𝑥+1)(𝑥−1)(𝑥2+1)
1 = 𝐴(𝑥 − 1)(𝑥2
+ 1) + 𝐵(𝑥 + 1)(𝑥2
+ 1) + (𝐶𝑥 + 𝐷) (𝑥 + 1)(𝑥 − 1)
1 = 𝐴(𝑥3
− 𝑥2
+ 𝑥 − 1) + 𝐵(𝑥3
+ 𝑥2
+ 𝑥 + 1) + 𝐶 (𝑥3
− 𝑥) + 𝐷(𝑥2
− 1) -----(b)
For, 𝑥 = 1, we have
𝐵 =
1
4

35. For, 𝑥 = −1, we have
𝐴 = −
1
4
For, 𝑥 = 0, we have
1 = −𝐴 + 𝐵 − 𝐷
𝐷 =
1
4
+
1
4
− 1
𝐷 =
1
2
− 1
𝐷 = −
1
2
Equating the coefficients of 𝑥3
on both sides of (b), we get
0 = 𝐴 + 𝐵 + 𝐶
0 = −
1
4
+
1
4
+ 𝐶
 𝐶 = 0
Now,
𝐼 = ∫
𝑑𝑥
(𝑥+1)(𝑥−1)(𝑥2+1)
𝐼 = ∫ (
𝐴
(𝑥+1)
+
𝐵
(𝑥−1)
+
𝐶𝑥+𝐷
(𝑥2+1)
) 𝑑𝑥 Using (a)
𝐼 = ∫
𝐴
(𝑥+1)
𝑑𝑥 + ∫
𝐵
(𝑥−1)
𝑑𝑥 + ∫
𝐶𝑥+𝐷
(𝑥2+1)
𝑑𝑥
𝐼 = 𝐴 ∫
𝑑𝑥
(𝑥+1)
+ 𝐵 ∫
𝑑𝑥
(𝑥−1)
+ ∫
𝐷
(𝑥2+1)
𝑑𝑥 Since,𝐶 = 0
𝐼 = 𝐴 ∫
𝑑𝑥
(𝑥+1)
+ 𝐵 ∫
𝑑𝑥
(𝑥−1)
+ 𝐷 ∫
𝑑𝑥
(𝑥2+1)
𝐼 = 𝐴 log(𝑥 + 1) + 𝐵 log(𝑥 − 1) + 𝐷 tan−1
𝑥
𝐼 = −
1
4
log(𝑥 + 1) +
1
4
log(𝑥 − 1) −
1
2
tan−1
𝑥
𝐼 =
1
4
log(𝑥 − 1) −
1
4
log(𝑥 + 1) −
1
2
tan−1
𝑥
𝐼 =
1
4
{log(𝑥 − 1) − log(𝑥 + 1)} −
1
2
tan−1
𝑥
𝐼 =
1
4
log
𝑥−1
𝑥+1
−
1
2
tan−1
𝑥
17. (i) ∫
𝑥 𝑑𝑥
(𝑥2+𝑎2)(𝑥2+𝑏2)
(ii) ∫
𝑥2 𝑑𝑥
(𝑥2+𝑎2)(𝑥2+𝑏2)

36. (i) Solution: Let , 𝐼 = ∫
𝑥 𝑑𝑥
(𝑥2+𝑎2)(𝑥2+𝑏2)
Take, 𝑥2
= 𝑢  𝑥 𝑑𝑥 =
1
2
𝑑𝑢
𝐼 = ∫
1
2
𝑑𝑢
(𝑢+𝑎2)(𝑢+𝑏2)
𝐼 =
1
2
∫
𝑑𝑢
(𝑢+𝑎2)(𝑢+𝑏2)
Let,
1
(𝑢+𝑎2)(𝑢+𝑏2)
=
𝐴
𝑢+𝑎2 +
𝐵
𝑢+𝑏2 ------- (a)
1 = 𝐴(𝑢 + 𝑏2) + 𝐵(𝑢 + 𝑎2)
Putting, 𝑢 = −𝑎2
in (b), we get
𝐴 =
1
𝑏2−𝑎2 = −
1
𝑎2−𝑏2
Again, putting 𝑢 = −𝑏2
in (b), we get
𝐵 ==
1
𝑎2−𝑏2
Now,
𝐼 =
1
2
∫
𝑑𝑢
(𝑢+𝑎2)(𝑢+𝑏2)
𝐼 =
1
2
∫ (
𝐴
𝑢+𝑎2 +
𝐵
𝑢+𝑏2
) 𝑑𝑢 Using (a)
𝐼 =
1
2
∫
𝐴
𝑢+𝑎2 𝑑𝑢 +
1
2
∫
𝐵
𝑢+𝑏2 𝑑𝑢
𝐼 =
1
2
𝐴 ∫
𝑑𝑢
𝑢+𝑎2 +
1
2
𝐵 ∫
𝑑𝑢
𝑢+𝑏2
𝐼 =
1
2
𝐴 log(𝑢 + 𝑎2) +
1
2
𝐵 log(𝑢 + 𝑏2)
𝐼 =
1
2
(−
1
𝑎2−𝑏2
) log(𝑢 + 𝑎2) +
1
2
(
1
𝑎2−𝑏2
)log(𝑢 + 𝑏2)
𝐼 = −
1
2(𝑎2−𝑏2)
log(𝑢 + 𝑎2) +
1
2(𝑎2−𝑏2)
log(𝑢 + 𝑏2)
𝐼 =
1
2(𝑎2−𝑏2)
{log(𝑢 + 𝑏2) − log(𝑢 + 𝑎2)}
𝐼 =
1
2(𝑎2−𝑏2)
{log
𝑢+𝑏2
𝑢+𝑎2
}
𝐼 =
1
2(𝑎2−𝑏2)
log
𝑥2+𝑏2
𝑥2+𝑏2

37. (ii) Solution: Let 𝐼 = ∫
𝑥2 𝑑𝑥
(𝑥2+𝑎2)(𝑥2+𝑏2)
Substituting, 𝑥2
= 𝑢 , for a partial fraction, we have
𝑥2
(𝑥2+𝑎2)(𝑥2+𝑏2)
=
𝑢
(𝑢+𝑎2)(𝑢+𝑏2)
Here,
𝑢
(𝑢+𝑎2)(𝑢+𝑏2)
=
𝐴
(𝑢+𝑎2)
+
𝐵
(𝑢+𝑏2)
------- (a)
𝑢 = 𝐴(𝑢 + 𝑏2) + 𝐵(𝑢 + 𝑎2) ------ (b)
Putting, 𝑢 = −𝑎2
in (b), we get
𝐴 =
𝑎2
𝑎2−𝑏2
Putting, 𝑢 = −𝑏2
in (b), we get
𝐵 = −
𝑏2
𝑎2−𝑏2
Now,
𝐼 = ∫
𝑥2 𝑑𝑥
(𝑥2+𝑎2)(𝑥2+𝑏2)
𝐼 = ∫ (
𝐴
(𝑢+𝑎2)
+
𝐵
(𝑢+𝑏2)
) 𝑑𝑥 Using (a)
𝐼 = ∫ (
𝐴
(𝑥2+𝑎2)
+
𝐵
(𝑥2+𝑏2)
) 𝑑𝑥 Since, 𝑥2
= 𝑢
𝐼 = ∫
𝐴
𝑥2+𝑎2 𝑑𝑥 + ∫
𝐵
𝑥2+𝑏2 𝑑𝑥
𝐼 = 𝐴 ∫
𝑑𝑥
𝑥2+𝑎2 + 𝐵 ∫
𝑑𝑥
𝑥2+𝑏2
𝐼 = 𝐴
1
𝑎
tan−1 𝑥
𝑎
+ 𝐵
1
𝑏
tan−1 𝑥
𝑏
𝐼 =
𝑎2
𝑎2−𝑏2 
1
𝑎
tan−1 𝑥
𝑎
−
𝑏2
𝑎2−𝑏2 
1
𝑏
tan−1 𝑥
𝑏
𝐼 =
𝑎
𝑎2−𝑏2 tan−1 𝑥
𝑎
−
𝑏
𝑎2−𝑏2 tan−1 𝑥
𝑏
𝐼 =
1
𝑎2−𝑏2
{𝑎 tan−1 𝑥
𝑎
− 𝑏 tan−1 𝑥
𝑏
}
18. (i) ∫
𝑥3 𝑑𝑥
(𝑥2+𝑎2)(𝑥2+𝑏2)
(ii) ∫
𝑥4 𝑑𝑥
(𝑥2+𝑎2)(𝑥2+𝑏2)
(i) Solution: Let 𝐼 = ∫
𝑥3 𝑑𝑥
(𝑥2+𝑎2)(𝑥2+𝑏2)

38. Let,
𝑥3
(𝑥2+𝑎2)(𝑥2+𝑏2)
=
𝐴𝑥
(𝑥2+𝑎2)
+
𝐵𝑥
(𝑥2+𝑏2)
----- (a)
𝑥3
(𝑥2+𝑎2)(𝑥2+𝑏2)
=
𝐴𝑥(𝑥2+𝑏2)+𝐵𝑥(𝑥2+𝑎2)
(𝑥2+𝑎2)(𝑥2+𝑏2)
𝑥3
= 𝐴𝑥(𝑥2
+ 𝑏2) + 𝐵𝑥(𝑥2
+ 𝑎2)
𝑥3
= 𝐴(𝑥3
+ 𝑏2
𝑥) + 𝐵(𝑥3
+ 𝑎2
𝑥) ------- (b)
Equating the coefficients of 𝑥3
on both sides of (b), we get
1 = 𝐴 + 𝐵
 𝐴 = −1 − 𝐵 ------- (i)
Equating the coefficients of 𝑥 on both sides of (b), we get
0 = 𝐴𝑏2
+ 𝐵𝑎2
𝐵𝑎2
= −𝐴𝑏2
𝐵 = −𝐴
𝑏2
𝑎2 ------ (ii)
Putting, 𝐵 = −𝐴
𝑏2
𝑎2 in (i) , we get
𝐴 = −1 − 𝐵
𝐴 = −1 + 𝐴
𝑏2
𝑎2
𝐴 − 𝐴
𝑏2
𝑎2 = −1
𝐴 (1 −
𝑏2
𝑎2
) = −1
𝐴 (
𝑎2−𝑏2
𝑎2
) = −1
𝐴 = −
𝑎2
𝑎2−𝑏2
𝐴 =
𝑎2
𝑏2−𝑎2
Again, Putting 𝐴 =
𝑎2
𝑏2−𝑎2 , in (ii), we get
𝐵 = −𝐴
𝑏2
𝑎2
𝐵 = −
𝑎2
𝑏2−𝑎2 
𝑏2
𝑎2 = −
𝑏2
𝑏2−𝑎2
Now,

39. 𝐼 = ∫
𝑥3 𝑑𝑥
(𝑥2+𝑎2)(𝑥2+𝑏2)
𝐼 = ∫ (
𝐴𝑥
(𝑥2+𝑎2)
+
𝐵𝑥
(𝑥2+𝑏2)
) 𝑑𝑥 Using (a)
𝐼 = ∫
𝐴𝑥
(𝑥2+𝑎2)
𝑑𝑥 + ∫
𝐵𝑥
(𝑥2+𝑏2)
𝑑𝑥
𝐼 = 𝐴 ∫
𝑥 𝑑𝑥
(𝑥2+𝑎2)
+ 𝐵 ∫
𝑥 𝑑𝑥
(𝑥2+𝑎2)
Take, 𝑥2
+ 𝑎2
= 𝑢 and 𝑥2
+ 𝑎2
= 𝑣
 𝑥 𝑑𝑥 =
1
2
𝑑𝑢 and 𝑥 𝑑𝑥 =
1
2
𝑑𝑣
𝐼 = 𝐴 ∫
1
2
𝑑𝑢
𝑢
+ 𝐵 ∫
1
2
𝑑𝑣
𝑣
𝐼 =
1
2
𝐴 ∫
𝑑𝑢
𝑢
+
1
2
𝐵 ∫
𝑑𝑣
𝑣
𝐼 =
1
2
𝐴 log 𝑢 +
1
2
𝐵 log 𝑣
𝐼 =
1
2
𝐴 log(𝑥2
+ 𝑎2) +
1
2
𝐵 log(𝑥2
+ 𝑎2)
𝐼 =
1
2

𝑎2
𝑏2−𝑎2 log(𝑥2
+ 𝑎2) +
1
2
 (−
𝑏2
𝑏2−𝑎2
) log(𝑥2
+ 𝑎2)
𝐼 =
1
2(𝑏2−𝑎2)
{log(𝑥2
+ 𝑎2) − 𝑏2
log(𝑥2
+ 𝑎2) }
(ii) Solution: Let 𝐼 = ∫
𝑥4 𝑑𝑥
(𝑥2+𝑎2)(𝑥2+𝑏2)
𝐼 = ∫
𝑥4
𝑥4+ 𝑏2𝑥2+𝑎2𝑥2+𝑎2𝑏2 𝑑𝑥
𝐼 = ∫ {1 − (
𝑏2𝑥2+𝑎2𝑥2+𝑎2𝑏2
𝑥4+ 𝑏2𝑥2+𝑎2𝑥2+𝑎2𝑏2
)} 𝑑𝑥 On dividing the numerator by denominator.
𝐼 = ∫ 1 𝑑𝑥 − ∫
𝑏2𝑥2+𝑎2𝑥2+𝑎2𝑏2
𝑥4+ 𝑏2𝑥2+𝑎2𝑥2+𝑎2𝑏2 𝑑𝑥
𝐼 = 𝑥 − ∫
𝑎2(𝑥2+𝑏2)+𝑏2𝑥2
(𝑥2+𝑎2)(𝑥2+𝑏2)
𝑑𝑥
𝐼 = 𝑥 − {∫
𝑎2(𝑥2+𝑏2)
(𝑥2+𝑎2)(𝑥2+𝑏2)
𝑑𝑥 + ∫
𝑏2𝑥2
(𝑥2+𝑎2)(𝑥2+𝑏2)
𝑑𝑥}
𝐼 = 𝑥 − {∫
𝑎2
(𝑥2+𝑎2)
𝑑𝑥 + 𝑏2
∫
𝑥2
(𝑥2+𝑎2)(𝑥2+𝑏2)
𝑑𝑥}
𝐼 = 𝑥 − {𝑎2
∫
𝑑𝑥
(𝑥2+𝑎2)
+ 𝑏2
∫
𝑥2
(𝑥2+𝑎2)(𝑥2+𝑏2)
𝑑𝑥}
𝐼 = 𝑥 − {𝑎2

1
𝑎
tan−1 𝑥
𝑎
+ 𝑏2
∫
𝑥2
(𝑥2+𝑎2)(𝑥2+𝑏2)
𝑑𝑥}

40. 𝐼 = 𝑥 − {𝑎 tan−1 𝑥
𝑎
+ 𝑏2
∫
𝑥2
(𝑥2+𝑎2)(𝑥2+𝑏2)
𝑑𝑥}
𝐼 = 𝑥 − 𝑎 tan−1 𝑥
𝑎
− 𝑏2
𝐼1 Where, 𝐼1 = ∫
𝑥2
(𝑥2+𝑎2)(𝑥2+𝑏2)
𝑑𝑥
Here,
𝐼1 = ∫
𝑥2
(𝑥2+𝑎2)(𝑥2+𝑏2)
𝑑𝑥
Substituting, 𝑥2
= 𝑢 , for a partial fraction, we have
𝑥2
(𝑥2+𝑎2)(𝑥2+𝑏2)
=
𝑢
(𝑢+𝑎2)(𝑢+𝑏2)
Here,
𝑢
(𝑢+𝑎2)(𝑢+𝑏2)
=
𝐴
(𝑢+𝑎2)
+
𝐵
(𝑢+𝑏2)
------- (a)
𝑢 = 𝐴(𝑢 + 𝑏2) + 𝐵(𝑢 + 𝑎2) ------ (b)
Putting, 𝑢 = −𝑎2
in (b), we get
𝐴 =
𝑎2
𝑎2−𝑏2
Putting, 𝑢 = −𝑏2
in (b), we get
𝐵 == −
𝑏2
𝑎2−𝑏2
Now,
𝐼1 = ∫
𝑥2 𝑑𝑥
(𝑥2+𝑎2)(𝑥2+𝑏2)
𝐼1 = ∫ (
𝐴
(𝑢+𝑎2)
+
𝐵
(𝑢+𝑏2)
) 𝑑𝑥 Using (a)
𝐼1 = ∫ (
𝐴
(𝑥2+𝑎2)
+
𝐵
(𝑥2+𝑏2)
) 𝑑𝑥 Since, 𝑥2
= 𝑢
𝐼1 = ∫
𝐴
𝑥2+𝑎2 𝑑𝑥 + ∫
𝐵
𝑥2+𝑏2 𝑑𝑥
𝐼1 = 𝐴 ∫
𝑑𝑥
𝑥2+𝑎2 + 𝐵 ∫
𝑑𝑥
𝑥2+𝑏2
𝐼1 = 𝐴
1
𝑎
tan−1 𝑥
𝑎
+ 𝐵
1
𝑏
tan−1 𝑥
𝑏
𝐼1 =
𝑎2
𝑎2−𝑏2 
1
𝑎
tan−1 𝑥
𝑎
−
𝑏2
𝑎2−𝑏2 
1
𝑏
tan−1 𝑥
𝑏
𝐼1 =
𝑎
𝑎2−𝑏2 tan−1 𝑥
𝑎
−
𝑏
𝑎2−𝑏2 tan−1 𝑥
𝑏
𝐼1 =
1
𝑎2−𝑏2
{𝑎 tan−1 𝑥
𝑎
− 𝑏 tan−1 𝑥
𝑏
}

41. 𝑇ℎ𝑢𝑠,
𝐼 = 𝑥 − 𝑎 tan−1 𝑥
𝑎
− 𝑏2
𝐼1
𝐼 = 𝑥 − 𝑎 tan−1 𝑥
𝑎
− 𝑏2 (
1
𝑎2−𝑏2
{𝑎 tan−1 𝑥
𝑎
− 𝑏 tan−1 𝑥
𝑏
})
𝐼 = 𝑥 − 𝑎 tan−1 𝑥
𝑎
−
𝑎𝑏2
𝑎2−𝑏2 tan−1 𝑥
𝑎
+
𝑏3
𝑎2−𝑏2 tan−1 𝑥
𝑏
𝐼 = 𝑥 − tan−1 𝑥
𝑎
(𝑎 +
𝑎𝑏2
𝑎2−𝑏2
) +
𝑏3
𝑎2−𝑏2 tan−1 𝑥
𝑏
𝐼 = 𝑥 − tan−1 𝑥
𝑎
(
𝑎3−𝑎𝑏2∓𝑎𝑏2
𝑎2−𝑏2
) +
𝑏3
𝑎2−𝑏2 tan−1 𝑥
𝑏
𝐼 = 𝑥 − tan−1 𝑥
𝑎
(
𝑎3
𝑎2−𝑏2
) +
𝑏3
𝑎2−𝑏2 tan−1 𝑥
𝑏
𝐼 = 𝑥 − (
𝑎3
𝑎2−𝑏2
)tan−1 𝑥
𝑎
+
𝑏3
𝑎2−𝑏2 tan−1 𝑥
𝑏
𝐼 = 𝑥 −
𝑎3
−(𝑏2−𝑎2)
tan−1 𝑥
𝑎
+
𝑏3
𝑎2−𝑏2 tan−1 𝑥
𝑏
𝐼 = 𝑥 +
𝑎3
𝑏2−𝑎2 tan−1 𝑥
𝑎
+
𝑏3
𝑎2−𝑏2 tan−1 𝑥
𝑏
19. ∫
𝑑𝑥
(𝑥2 + 𝑎2)(𝑥+𝑏)
Solution: Let 𝐼 = ∫
𝑑𝑥
(𝑥2 + 𝑎2)(𝑥+𝑏)
Let,
1
(𝑥2 + 𝑎2)(𝑥+𝑏)
=
𝐴𝑥+𝐵
(𝑥2 + 𝑎2)
+
𝐶
(𝑥+𝑏)
------ (a)
1
(𝑥2 + 𝑎2)(𝑥+𝑏)
=
(𝐴𝑥+𝐵)(𝑥+𝑏)+ 𝐶 (𝑥2 + 𝑎2)
(𝑥2 + 𝑎2)
1 = (𝐴𝑥 + 𝐵)(𝑥 + 𝑏) + 𝐶 (𝑥2
+ 𝑎2)
1 = 𝐴(𝑥2
+ 𝑏 𝑥) + 𝐵(𝑥 + 𝑏) + 𝐶 (𝑥2
+ 𝑎2) -------- (b)
Equating the coefficients of 𝑥2
on both sides of equation (b), we get
𝑂 = 𝐴 + 𝐶  𝐴 = −𝐶 ----- (i)
Equating the coefficients of 𝑥2
on both sides of equation (b), we get
𝑂 = 𝑏𝐴 + 𝐵  𝐵 = −𝑏𝐴 ------ (ii)
Equating the constant terms on both sides of equation (b), we get
1 = 𝑏𝐵 + 𝑎2
𝐶 -------- (iii)
Putting 𝐴 = −𝐶 in (ii), we get

42. 𝐵 = 𝑏𝐶
Again, putting 𝐵 = 𝑏𝐶 in (iii)
1 = 𝑏𝐵 + 𝑎2
𝐶
1 = 𝑏𝑏𝐶 + 𝑎2
𝐶
1 = 𝑏2
𝐶 + 𝑎2
𝐶
1 = 𝐶(𝑎2
+ 𝑏2)
 𝐶 =
1
𝑎2+𝑏2
Putting, 𝐶 =
1
𝑎2+𝑏2 in (i)
𝐴 = −𝐶  𝐴 = −
1
𝑎2+𝑏2
Finally, putting 𝐴 = −
1
𝑎2+𝑏2 in (ii) , we get
𝐵 = −𝑏𝐴  𝐵 =
𝑏
𝑎2+𝑏2
Now,
𝐼 = ∫
𝑑𝑥
(𝑥2 + 𝑎2)(𝑥+𝑏)
𝐼 = ∫ {
𝐴𝑥+𝐵
(𝑥2 + 𝑎2)
+
𝐶
(𝑥+𝑏)
} 𝑑𝑥
𝐼 = ∫
𝐴𝑥+𝐵
(𝑥2 + 𝑎2)
𝑑𝑥 + ∫
𝐶
(𝑥+𝑏)
𝑑𝑥
𝐼 = ∫
𝐴𝑥
(𝑥2 + 𝑎2)
𝑑𝑥 + 𝐵 ∫
𝑑𝑥
(𝑥2 + 𝑎2)
+ 𝐶 ∫
𝑑𝑥
(𝑥+𝑏)
𝐼 = 𝐴 ∫
𝑥
(𝑥2 + 𝑎2)
𝑑𝑥 + 𝐵
𝑥
𝑎
tan−1 𝑥
𝑎
+ 𝐶 log(𝑥 + 𝑏)
Putting, 𝑥2
+ 𝑎2
= 𝑢  𝑥 𝑑𝑥 =
1
2
𝑑𝑢
𝐼 = 𝐴 ∫
𝑥
(𝑥2 + 𝑎2)
𝑑𝑥 + 𝐵
𝑥
𝑎
tan−1 𝑥
𝑎
+ 𝐶 log(𝑥 + 𝑏)
𝐼 = 𝐴 ∫
1
2
𝑑𝑢
𝑢
+ 𝐵
𝑥
𝑎
tan−1 𝑥
𝑎
+ 𝐶 log(𝑥 + 𝑏)
𝐼 =
1
2
𝐴 ∫
𝑑𝑢
𝑢
+ 𝐵
𝑥
𝑎
tan−1 𝑥
𝑎
+ 𝐶 log(𝑥 + 𝑏)
𝐼 =
1
2
𝐴 log 𝑢 + 𝐵
𝑥
𝑎
tan−1 𝑥
𝑎
+ 𝐶 log(𝑥 + 𝑏)
𝐼 = 𝐴 log 𝑢
1
2 + 𝐵
𝑥
𝑎
tan−1 𝑥
𝑎
+ 𝐶 log(𝑥 + 𝑏)

43. 𝐼 = 𝐴 log √𝑢 + 𝐵
𝑥
𝑎
tan−1 𝑥
𝑎
+ 𝐶 log(𝑥 + 𝑏)
𝐼 = −
1
𝑎2+𝑏2 log √𝑥2 + 𝑎2 +
𝑏
𝑎2+𝑏2
1
𝑎
tan−1 𝑥
𝑎
+
𝑏
𝑎2+𝑏2 log(𝑥 + 𝑏)
𝐼 =
1
𝑎2+𝑏2
(log(𝑥 + 𝑏) − log √𝑥2 + 𝑎2) +
𝑏
𝑎2+𝑏2
1
𝑎
tan−1 𝑥
𝑎
𝐼 =
1
𝑎2+𝑏2 log
(𝑥+𝑏)
√𝑥2 + 𝑎2
+
𝑏
𝑎2+𝑏2
1
𝑎
tan−1 𝑥
𝑎
𝐼 =
1
𝑎2+𝑏2
{log
(𝑥+𝑏)
√𝑥2 + 𝑎2
+
𝑏
𝑎
tan−1 𝑥
𝑎
}
20. (i) ∫
𝑥 𝑑𝑥
(1+𝑥)(1+𝑥2)
(ii) ∫
𝑥
𝑥3−1
𝑑𝑥
(i) Solution: Let 𝐼 = ∫
𝑥 𝑑𝑥
(1+𝑥)(1+𝑥2)
Let,
𝑥
(1+𝑥)(1+𝑥2)
=
𝐴
(1+𝑥)
+
𝐵𝑥+𝐶
(1+𝑥2)
-------- (a)
𝑥
(1+𝑥)(1+𝑥2)
=
𝐴(1+𝑥2)+ (𝐵𝑥+𝐶)(1+𝑥)
(1+𝑥)(1+𝑥2)
𝑥 = 𝐴(1 + 𝑥2) + 𝐵(𝑥 + 𝑥2) + 𝐶(1 + 𝑥) -------- (b)
For, 𝑥 = 0, we get
0 = 𝐴 + 𝐶 𝐶 = −𝐴 -------- (i)
For, 𝑥 = −1, we get
−1 = 2𝐴  𝐴 = −
1
2
Putting, 𝐴 = −
1
2
in (i), we get
𝐶 = −𝐴  𝐶 =
1
2
Equating the coefficients of 𝑥2
in both sides of equation (b), we get
0 = 𝐴 + 𝐵
𝐵 = −𝐴
𝐵 =
1
2
Now,
𝐼 = ∫
𝑥 𝑑𝑥
(1+𝑥)(1+𝑥2)
𝐼 = ∫ (
𝐴
(1+𝑥)
+
𝐵𝑥+𝐶
(1+𝑥2)
) 𝑑𝑥 Using (a)

44. 𝐼 = ∫
𝐴
(1+𝑥)
𝑑𝑥 + ∫
𝐵𝑥
(1+𝑥2)
𝑑𝑥 + ∫
𝐶
(1+𝑥2)
𝑑𝑥
𝐼 = 𝐴 ∫
𝑑𝑥
(1+𝑥)
+ 𝐵 ∫
𝑥 𝑑𝑥
(1+𝑥2)
+ 𝐶 ∫
𝑑𝑥
(1+𝑥2)
𝐼 = 𝐴 log(1 + 𝑥) + 𝐵 ∫
𝑥 𝑑𝑥
(1+𝑥2)
+ 𝐶 tan−1
𝑥
𝐼 = 𝐴 log(1 + 𝑥) + 𝐵 log(1 + 𝑥2) + 𝐶 tan−1
𝑥
𝐼 = −
1
2
log(1 + 𝑥) +
1
2

1
2
log(1 + 𝑥2) +
1
2
tan−1
𝑥
𝐼 = −
1
2
log(1 + 𝑥) +
1
4
log(1 + 𝑥2) +
1
2
tan−1
𝑥
(ii) Solution: Let 𝐼 = ∫
𝑥
𝑥3−1
𝑑𝑥
𝐼 = ∫
𝑥
(𝑥−1)(𝑥2+𝑥+1)
𝑑𝑥
Let,
𝑥
(𝑥−1)(𝑥2+𝑥+1)
=
𝐴
(𝑥−1)
+
𝐵𝑥+𝐶
(𝑥2+𝑥+1)
------ (a)
𝑥
(𝑥−1)(𝑥2+𝑥+1)
=
𝐴(𝑥2+𝑥+1)+(𝐵𝑥+𝐶)(𝑥−1)
(𝑥−1)(𝑥2+𝑥+1)
𝑥 = 𝐴(𝑥2
+ 𝑥 + 1) + (𝐵𝑥 + 𝐶)(𝑥 − 1)
𝑥 = 𝐴(𝑥2
+ 𝑥 + 1) + 𝐵(𝑥2
− 𝑥) + 𝐶(𝑥 − 1) ---------- (b)
For, = 1 , we have
1 = 3𝐴  𝐴 =
1
3
For, = 0 , we have
0 = 𝐴 − 𝐶
𝐶 = 𝐴
 𝐶 =
1
3
Equating the coefficients of 𝑥2
, on both sides of (b)
0 = 𝐴 + 𝐵
𝐵 = −𝐴
𝐵 = −
1
3
Now,
𝐼 = ∫
𝑥
(𝑥−1)(𝑥2+𝑥+1)
𝑑𝑥

45. 𝐼 = ∫ {
𝐴
(𝑥−1)
+
𝐵𝑥+𝐶
(𝑥2+𝑥+1)
} 𝑑𝑥
𝐼 = ∫
𝐴
(𝑥−1)
𝑑𝑥 + ∫
𝐵𝑥+𝐶
(𝑥2+𝑥+1)
𝑑𝑥
𝐼 =
1
3
∫
𝑑𝑥
(𝑥−1)
+ ∫
−
1
3
𝑥+
1
3
(𝑥2+𝑥+1)
𝑑𝑥
𝐼 =
1
3
∫
𝑑𝑥
(𝑥−1)
−
1
3
∫
𝑥−1
(𝑥2+𝑥+1)
𝑑𝑥
𝐼 =
1
3
∫
𝑑𝑥
(𝑥−1)
−
1
6
∫
2𝑥−2
(𝑥2+𝑥+1)
𝑑𝑥
𝐼 =
1
3
∫
𝑑𝑥
(𝑥−1)
−
1
6
∫
2𝑥+1−3
(𝑥2+𝑥+1)
𝑑𝑥 +
1
6
∫
3
(𝑥2+𝑥+1)
𝑑𝑥
𝐼 =
1
3
∫
𝑑𝑥
(𝑥−1)
−
1
6
∫
2𝑥+1
(𝑥2+𝑥+1)
𝑑𝑥 +
3
6
∫
𝑑𝑥
(𝑥2+𝑥+1)
𝐼 =
1
3
log(𝑥 − 1) −
1
6
∫
2𝑥+1
(𝑥2+𝑥+1)
𝑑𝑥 +
1
2
∫
𝑑𝑥
(𝑥+
1
2
)
2
+(
√3
2
)
2
Putting, 𝑥2
+ 𝑥 + 1 = 𝑢  (2𝑥 + 1)𝑑𝑥 = 𝑑𝑢
𝐼 =
1
3
log(𝑥 − 1) −
1
6
∫
𝑑𝑢
𝑢
+
1
2
∫
𝑑𝑥
(𝑥+
1
2
)
2
+(
√3
2
)
2
𝐼 =
1
3
log(𝑥 − 1) −
1
6
log 𝑢 +
1
2

1
√3
2
tan−1
(
𝑥+
1
2
√3
2
)
𝐼 =
1
3
log(𝑥 − 1) −
1
6
log(𝑥2
+ 𝑥 + 1) +
1
√3
tan−1 (
2𝑥+1
√3
)
𝐼 =
1
3
log(𝑥 − 1) −
1
3

1
2
log(𝑥2
+ 𝑥 + 1) +
1
√3
tan−1 (
2𝑥+1
√3
)
𝐼 =
1
3
log(𝑥 − 1) −
1
3
log(𝑥2
+ 𝑥 + 1)
1
2 +
1
√3
tan−1 (
2𝑥+1
√3
)
𝐼 =
1
3
log(𝑥 − 1) −
1
3
log √𝑥2 + 𝑥 + 1 +
1
√3
tan−1 (
2𝑥+1
√3
)
𝐼 =
1
3
(log(𝑥 − 1) − log √𝑥2 + 𝑥 + 1) +
1
√3
tan−1 (
2𝑥+1
√3
)
𝐼 =
1
3
log
𝑥−1
√𝑥2+𝑥+1
+
1
√3
tan−1 (
2𝑥+1
√3
)
21. ∫
(𝑥2−1)
𝑥4+𝑥2+1
𝑑𝑥
Solution: 𝐼 = ∫
(𝑥2−1)
𝑥4+𝑥2+1
𝑑𝑥

46. 𝐼 = ∫
𝑥2(1−
1
𝑥2)
𝑥2(𝑥2+1+
1
𝑥2)
𝑑𝑥
𝐼 = ∫
(1−
1
𝑥2)
(𝑥2+1+
1
𝑥2)
𝑑𝑥
𝐼 = ∫
(1−
1
𝑥2)
(𝑥+
1
𝑥
)
2
−1
𝑑𝑥
Putting, 𝑥 +
1
𝑥
= 𝑢  (1 −
1
𝑥2
) 𝑑𝑥 = 𝑑𝑢
𝐼 = ∫
𝑑𝑢
𝑢2−1
𝐼 = ∫
𝑑𝑢
(𝑢+1)(𝑢−1)
Let,
1
(𝑢+1)(𝑢−1)
=
𝐴
(𝑢+1)
+
𝐵
(𝑢−1)
------ (a)
1
(𝑢+1)(𝑢−1)
=
𝐴(𝑢−1)+ 𝐵 (𝑢+1)
(𝑢+1)
1 = 𝐴(𝑢 − 1) + 𝐵 (𝑢 + 1) ------ (b)
Putting, 𝑢 = 1 and 𝑢 = −1 in (b) , we get
𝐴 = −
1
2
and 𝐵 =
1
2
Now,
𝐼 = ∫
𝑑𝑢
(𝑢+1)(𝑢−1)
𝐼 = ∫ (
𝐴
(𝑢+1)
+
𝐵
(𝑢−1)
) 𝑑𝑢
𝐼 = ∫
𝐴
(𝑢+1)
𝑑𝑢 + ∫
𝐵
(𝑢−1)
𝑑𝑢
𝐼 = 𝐴 ∫
𝑑𝑢
(𝑢+1)
+ 𝐵 ∫
𝑑𝑢
(𝑢−1)
𝐼 = 𝐴 log(𝑢 + 1) + 𝐵 log(𝑢 − 1)
𝐼 = −
1
2
log(𝑢 + 1) +
1
2
log(𝑢 − 1)
𝐼 =
1
2
log(𝑢 − 1) −
1
2
log(𝑢 + 1)
𝐼 =
1
2
log (𝑥 +
1
𝑥
− 1) −
1
2
log (𝑥 +
1
𝑥
+ 1)
𝐼 =
1
2
log (
𝑥2−𝑥+1
𝑥
) −
1
2
log (
𝑥2+𝑥+1
𝑥
)
𝐼 =
1
2
log(𝑥2
− 𝑥 + 1) −
1
2
log 𝑥 −
1
2
log(𝑥2
+ 𝑥 + 1) +
1
2
log 𝑥

47. 𝐼 =
1
2
log(𝑥2
− 𝑥 + 1) −
1
2
log(𝑥2
+ 𝑥 + 1)
22. (i) ∫
𝑥2
𝑥4−𝑥2−12
𝑑𝑥 (ii) ∫
𝑥 𝑑𝑥
𝑥4−𝑥2−2
(i) Solution: Let, 𝐼 = ∫
𝑥2
𝑥4−𝑥2−12
𝑑𝑥
𝐼 = ∫
𝑥2
𝑥4−4𝑥2+3𝑥2−12
𝑑𝑥
𝐼 = ∫
𝑥2
𝑥2(𝑥2−4)+3(𝑥2−3)
𝑑𝑥
𝐼 = ∫
𝑥2
(𝑥2−4)(𝑥2+3)
𝑑𝑥
Substituting, 𝑥2
= 𝑢 for partial fraction
Here,
𝑥2
(𝑥2−4)(𝑥2+3)
=
𝑢
(𝑢−4)(𝑢+3)
Let,
𝑢
(𝑢−4)(𝑢+3)
=
𝐴
(𝑢−4)
+
𝐵
(𝑢+3)
------ (a)
𝑢
(𝑢−4)(𝑢+3)
=
𝐴(𝑢+3)+ 𝐵 (𝑢−4)
(𝑢−4)(𝑢+3)
𝑢 = 𝐴(𝑢 + 3) + 𝐵 (𝑢 − 4) ------- (b)
Putting, 𝑢 = −3 in (b), we get
𝐵 =
3
7
Putting, 𝑢 = 4 in (b), we get
𝐴 =
4
7
Now,
𝐼 = ∫
𝑥2
(𝑥2−4)(𝑥2+3)
𝑑𝑥
𝐼 = ∫ (
𝐴
(𝑢−4)
+
𝐵
(𝑢+3)
) 𝑑𝑥 Using (a)
𝐼 = ∫ (
𝐴
(𝑥2−4)
+
𝐵
(𝑥2+3)
) 𝑑𝑥
𝐼 = ∫
𝐴
(𝑥2−4)
𝑑𝑥 + ∫
𝐵
(𝑥2+3)
𝑑𝑥
𝐼 = 𝐴 ∫
𝑑𝑥
𝑥2−22 + 𝐵 ∫
𝑑𝑥
𝑥2+ (√3)
2

48. 𝐼 = 𝐴
1
22
log |
𝑥−2
𝑥+2
| + 𝐵
1
√3
tan−1 𝑥
√3
𝐼 =
4
7

1
4
log |
𝑥−2
𝑥+2
| +
3
7

1
√3
tan−1 𝑥
√3
𝐼 =
1
7
log |
𝑥−2
𝑥+2
| +
√3
7
tan−1 𝑥
√3
(ii) Solution: Let, 𝐼 = ∫
𝑥 𝑑𝑥
𝑥4−𝑥2−2
𝐼 = ∫
𝑥 𝑑𝑥
(𝑥2+1)(𝑥+ √2)(𝑥−√2)
Here,
𝑥
(𝑥2+1)(𝑥+ √2)(𝑥−√2)
=
𝐴𝑥+𝐵
(𝑥2+1)
+
𝐶
(𝑥+ √2)
+
𝐷
(𝑥−√2)
𝑥
(𝑥2+1)(𝑥+ √2)(𝑥−√2)
=
(𝐴𝑥+𝐵)(𝑥+ √2)(𝑥−√2)+ 𝐶(𝑥2+1)(𝑥−√2)+𝐷(𝑥2+1)(𝑥+ √2)
(𝑥2+1)(𝑥+ √2)(𝑥−√2)
𝑥 = (𝐴𝑥 + 𝐵)(𝑥 + √2)(𝑥 − √2) + 𝐶(𝑥2
+ 1)(𝑥 − √2) + 𝐷(𝑥2
+ 1)(𝑥 + √2) -------- (b)
Putting, = √2 , in (b) , we get
𝐷 =
√2
6√2
 𝐷 =
1
6
Putting, = −√2 , in (b) , we get
𝐶 =
−√2
6√2
 𝐶 =
1
6
Putting, = 0 , in (b) , we get
0 = 𝐵(−2) + 𝐶(−√2) + 𝐷(√2)
2𝐵 = 𝐶(−√2) + 𝐷(√2)
2𝐵 = −
1
6
√2 +
1
6
√2
2𝐵 = 0  𝐵 = 0
Equating the coefficients of 𝑥3
, on both sides of (b)
0 = 𝐴 + 𝐶 + 𝐷
𝐵 = −𝐶 − 𝐷
𝐵 = −
1
6
−
1
6
𝐵 = −
1
3
Now,
∫
𝑑𝑥
𝑥2−𝑎2 =
1
2𝑎
log |
𝑥−𝑎
𝑥+𝑎
|

49. 𝐼 = ∫
𝑥 𝑑𝑥
(𝑥2+1)(𝑥+ √2)(𝑥−√2)
𝐼 = ∫ (
𝐴𝑥+𝐵
(𝑥2+1)
+
𝐶
(𝑥+ √2)
+
𝐷
(𝑥−√2)
) 𝑑𝑥
𝐼 = ∫ (
𝐴𝑥
(𝑥2+1)
+
𝐶
(𝑥+ √2)
+
𝐷
(𝑥−√2)
) 𝑑𝑥 Since, 𝐵 = 0
𝐼 = ∫
𝐴𝑥
(𝑥2+1)
𝑑𝑥 + ∫
𝐶
(𝑥+ √2)
𝑑𝑥 + ∫
𝐷
(𝑥−√2)
𝑑𝑥
𝐼 = 𝐴 ∫
𝑥 𝑑𝑥
(𝑥2+1)
+ 𝐶 ∫
𝑑𝑥
(𝑥+ √2)
+ 𝐷 ∫
𝑑𝑥
(𝑥−√2)
𝐼 = 𝐴
1
2
log(𝑥2
+ 1) + 𝐶 log(𝑥 + √2) + 𝐷 log(𝑥 − √2)
𝐼 = −
1
3

1
2
log(𝑥2
+ 1) +
1
6
log(𝑥 + √2) +
1
6
log(𝑥 − √2)
𝐼 = −
1
6
log(𝑥2
+ 1) +
1
6
{log(𝑥 + √2) + log(𝑥 − √2)}
𝐼 = −
1
6
log(𝑥2
+ 1) +
1
6
{log(𝑥 + √2)(𝑥 − √2)}
𝐼 = −
1
6
log(𝑥2
+ 1) +
1
6
{log(𝑥2
− 4)}
𝐼 = −
1
6
log(𝑥2
+ 1) +
1
6
log(𝑥2
− 2)
𝐼 =
1
6
{log(𝑥2
− 2) − log(𝑥2
+ 1)}
23. ∫
𝑑𝑥
(𝑥4+ 𝑥2+1)2
Solution: Let 𝐼 = ∫
𝑑𝑥
(𝑥4+ 𝑥2+1)2
𝐼 = ∫
𝑑𝑥
{(𝑥+2)2+1}2
Take, 𝑥 + 2 = tan 𝜃  𝑑𝑥 = sec2
𝜃 𝑑𝜃
Now,
𝐼 = ∫
sec2 𝜃 𝑑𝜃
{tan2 𝜃+1}2
𝐼 = ∫
sec2 𝜃 𝑑𝜃
sec4 𝜃
𝐼 = ∫
𝑑𝜃
sec2 𝜃

50. 𝐼 = ∫ cos2
𝜃 𝑑𝜃
𝐼 = ∫
1
2
(1 + cos 2𝜃) 𝑑𝜃
𝐼 =
1
2
∫(1 + cos 2𝜃) 𝑑𝜃
𝐼 =
1
2
(𝜃 +
sin 2𝜃
2
)
𝐼 =
1
2
(𝜃 +
2sin 𝜃 cos𝜃
2
)
𝐼 =
1
2
(𝜃 + sin 𝜃 cos 𝜃)
𝐼 =
1
2
(tan−1(𝑥 + 2) +
𝑥+2
√𝑥2+4𝑥+5

1
√𝑥2+4𝑥+5
)
𝐼 =
1
2
(tan−1(𝑥 + 2) +
𝑥+2
𝑥2+4𝑥+5
)
24. ∫
𝑥2 𝑑𝑥
(𝑥2+1)(2𝑥2+1)
Solution: Let, 𝐼 = ∫
𝑥2 𝑑𝑥
(𝑥2+1)(2𝑥2+1)
𝑥2
(𝑥2+1)(2𝑥2+1)
=
𝐴𝑥+𝐵
(𝑥2+1)
+
𝐶𝑥+𝐷
(2𝑥2+1)
------ (a)
𝑥2
(𝑥2+1)(2𝑥2+1)
=
(𝐴𝑥+𝐵)(2𝑥2+1)+ (𝐶𝑥+𝐷)(𝑥2+1)
(𝑥2+1)(2𝑥2+1)
𝑥2
= (𝐴𝑥 + 𝐵)(2𝑥2
+ 1) + (𝐶𝑥 + 𝐷)(𝑥2
+ 1)
𝑥2
= 𝐴(2𝑥3
+ 𝑥) + 𝐵(2𝑥2
+ 1) + 𝐶(𝑥3
+ 𝑥) + 𝐷(𝑥2
+ 1) --------- (b)
Equating the coefficients of 𝑥3
on both sides of (b), we get
0 = 2𝐴 + 𝐶 𝐶 + 2𝐴 = 0 ------ (i)
Equating the coefficients of 𝑥2
on both sides of (b), we get
1 = 2𝐵 + 𝐷 ------ (ii)
Equating the coefficients of 𝑥 on both sides of (b), we get
0 = 𝐴 + 𝐶  𝐶 = −𝐴 ----- (iii)
We take,
𝑥 + 2 = tan 𝜃
𝜃 = tan−1(𝑥 + 2)
Again, 𝑥 + 2 = tan 𝜃
(𝑥 + 2)2
= tan2
𝜃
(𝑥 + 2)2
=
sin2 𝜃
cos2 𝜃
(𝑥 + 2)2
cos2
𝜃 = sin2
𝜃
(𝑥 + 2)2 (1 − sin2
𝜃) = sin2
𝜃
(𝑥 + 2)2
− (𝑥 + 2)2
sin2
𝜃 = sin2
𝜃
(𝑥 + 2)2
= sin2
𝜃 + (𝑥 + 2)2
sin2
𝜃
(𝑥 + 2)2
= sin2
𝜃 (1 + (𝑥 + 2)2)
(𝑥+2)2
(1+(𝑥+2)2)
= sin2
𝜃
sin 𝜃 =
𝑥+2
√1+(𝑥+2)2
sin 𝜃 =
𝑥+2
√𝑥2+4𝑥+5
and cos 𝜃 =
1
√𝑥2+4𝑥+5

51. Equating the constant terms on both sides of (b), we get
0 = 𝐵 + 𝐷  𝐷 = −𝐵 ----- (iv)
Putting, 𝐷 = −𝐵 in (ii), we get
1 = 2𝐵 + 𝐷
1 = 2𝐵 − 𝐵
 𝐵 = 1
Putting 𝐵 = 1 in (iv). We get
𝐷 = −𝐵  𝐷 = −1
Putting 𝐶 = −𝐴 in (i). We get
𝐶 + 2𝐴 = 0
−𝐴 + 2𝐴 = 0
𝐴 = 0
Putting 𝐴 = 0 in (i) , we get
𝐶 + 2𝐴 = 0
𝐶 + 0 = 0
𝐶 = 0
Now,
𝐼 = ∫
𝑥2 𝑑𝑥
(𝑥2+1)(2𝑥2+1)
𝐼 = ∫ (
𝐴𝑥+𝐵
(𝑥2+1)
+
𝐶𝑥+𝐷
(2𝑥2+1)
) 𝑑𝑥 Using (a)
𝐼 = ∫ (
𝐵
(𝑥2+1)
+
𝐷
(2𝑥2+1)
) 𝑑𝑥 Since, 𝐴 = 0 and 𝐶 = 0
𝐼 = ∫ (
1
(𝑥2+1)
+
−1
(2𝑥2+1)
) 𝑑𝑥
𝐼 = ∫ (
1
(𝑥2+1)
−
1
(2𝑥2+1)
) 𝑑𝑥
𝐼 = ∫
1
(𝑥2+1)
𝑑𝑥 − ∫
1
(2𝑥2+1)
𝑑𝑥
𝐼 = tan−1
𝑥 −
1
2
∫
𝑑𝑥
𝑥2+
1
2

52. 𝐼 = tan−1
𝑥 −
1
2
∫
𝑑𝑥
𝑥2+(
1
√2
)
2
𝐼 = tan−1
𝑥 −
1
2

1
1
√2
tan−1
(
𝑥
1
√2
)
𝐼 = tan−1
𝑥 −
√2
2
tan−1(√2 𝑥)
𝐼 = tan−1
𝑥 −
√2
2

√2
√2
tan−1(√2 𝑥)
𝐼 = tan−1
𝑥 −
2
2√2
tan−1(√2 𝑥)
𝐼 = tan−1
𝑥 −
1
√2
tan−1(√2 𝑥)
25. ∫
𝑑𝑥
𝑥(1+𝑥+𝑥2+𝑥3)
Solution: Let 𝐼 = ∫
𝑑𝑥
𝑥(1+𝑥+𝑥2+𝑥3)
𝐼 = ∫
𝑑𝑥
𝑥(𝑥+1)(𝑥2+1)
Let,
𝑑𝑥
𝑥(𝑥+1)(𝑥2+1)
=
𝐴
𝑥
+
𝐵
(𝑥+1)
+
𝐶𝑥+𝐷
(𝑥2+1)
---- (a)
1
𝑥(𝑥+1)(𝑥2+1)
=
𝐴(𝑥+1)(𝑥2+1)+ 𝐵𝑥(𝑥2+1)+(𝐶𝑥+𝐷)𝑥(𝑥+1)
𝑥 (𝑥+1)(𝑥2+1)
1 = 𝐴(𝑥3
+ 𝑥3
+ 𝑥2
+ 1) + 𝐵(𝑥3
+ 𝑥) + 𝐶(𝑥3
+ 𝑥2) + 𝐷(𝑥2
+ 𝑥) --- (b)
Putting, 𝑥 = −1 in (b), we get
1 = −2𝐵  𝐵 = −
1
2
Putting, 𝑥 = 0 in (b), we get
1 = 𝐴  𝐴 = 1
Equating the coefficients of 𝑥3
on both sides of (b), we get
0 = 𝐴 + 𝐵 + 𝐶
0 = 1 −
1
2
+ 𝐶
0 =
1
2
+ 𝐶
 𝐶 = −
1
2
Equating the coefficients of 𝑥2
on both sides of (b), we get

53. 0 = 𝐴 + 𝐶 + 𝐷
0 = 1 −
1
2
+ 𝐷
0 = −
1
2
+ 𝐷
𝐷 = −
1
2
Now,
𝐼 = ∫
𝑑𝑥
𝑥(𝑥+1)(𝑥2+1)
𝐼 = ∫ (
𝐴
𝑥
+
𝐵
(𝑥+1)
+
𝐶𝑥+𝐷
(𝑥2+1)
) 𝑑𝑥
𝐼 = ∫
𝐴
𝑥
𝑑𝑥 + ∫
𝐵
(𝑥+1)
𝑑𝑥 + ∫
𝐶𝑥+𝐷
(𝑥2+1)
𝑑𝑥
𝐼 = ∫
𝐴
𝑥
𝑑𝑥 + ∫
𝐵
(𝑥+1)
𝑑𝑥 + ∫
𝐶𝑥+𝐷
(𝑥2+1)
𝑑𝑥 + ∫
𝐷
(𝑥2+1)
𝑑𝑥
𝐼 = ∫
𝑑𝑥
𝑥
−
1
2
∫
𝑑𝑥
(𝑥+1)
+ ∫
−
1
2
𝑥
(𝑥2+1)
𝑑𝑥 + ∫
−
1
2
(𝑥2+1)
𝑑𝑥
𝐼 = log 𝑥 −
1
2
log(𝑥 + 1) −
1
2
∫
𝑥
(𝑥2+1)
𝑑𝑥 −
1
2
∫
𝑑𝑥
(𝑥2+1)
𝐼 = log 𝑥 −
1
2
log(𝑥 + 1) −
1
4
∫
2𝑥
(𝑥2+1)
𝑑𝑥 −
1
2
∫
𝑑𝑥
(𝑥2+1)
𝐼 = log 𝑥 −
1
2
log(𝑥 + 1) −
1
4
log(𝑥2
+ 1) −
1
2
tan−1
𝑥
26. ∫
𝑑𝑥
𝑥4+𝑥2+1
Solution: Let, 𝐼 = ∫
𝑑𝑥
𝑥4+𝑥2+1
𝐼 = ∫
𝑑𝑥
(𝑥2+𝑥+1)(𝑥2−𝑥+1)
Let,
1
(𝑥2+𝑥+1)(𝑥2−𝑥+1)
=
𝐴𝑥+𝐵
(𝑥2+𝑥+1)
+
𝐶𝑥+𝐷
(𝑥2−𝑥+1)
----- (a)
1
(𝑥2+𝑥+1)(𝑥2−𝑥+1)
=
(𝐴𝑥+𝐵)(𝑥2−𝑥+1)+(𝐶𝑥+𝐷) (𝑥2+𝑥+1)
(𝑥2+𝑥+1)(𝑥2−𝑥+1)
1 = (𝐴𝑥 + 𝐵)(𝑥2
− 𝑥 + 1) + (𝐶𝑥 + 𝐷) (𝑥2
+ 𝑥 + 1)
1 = 𝐴(𝑥3
− 𝑥2
+ 𝑥) + 𝐵(𝑥2
− 𝑥 + 1) + 𝐶 (𝑥3
+ 𝑥2
+ 𝑥) + 𝐷(𝑥2
+ 𝑥 + 1) ------- (b)
Equating the coefficients of 𝑥3
on both sides of equation (b), we get
0 = 𝐴 + 𝐶 𝐴 + 𝐶 = 0 ----- (i)

54. Equating the coefficients of 𝑥2
on both sides of equation (b), we get
0 = −𝐴 + 𝐵 + 𝐶 + 𝐷 ----- (ii)
Equating the coefficients of 𝑥 on both sides of equation (b), we get
0 = 𝐴 − 𝐵 + 𝐶 + 𝐷 -------- (iii)
Equating the constant terms on both sides of equation (b), we get
1 = 𝐵 + 𝐷 ------ (iv)
Adding (ii) and (iii), we get
0 = −𝐴 + 𝐵 + 𝐶 + 𝐷
0 = 𝐴 − 𝐵 + 𝐶 + 𝐷
------------------------------------------------
𝐶 + 𝐷 = 0 ----- (v)
Putting, 𝐴 + 𝐶 = 0 in (iii), we get
0 = 𝐴 − 𝐵 + 𝐶 + 𝐷
0 = −𝐵 + 𝐷
 𝐷 = 𝐵
Putting 𝐷 = 𝐵 in (iv)
1 = 𝐵 + 𝐷
1 = 𝐵 + 𝐵
1 = 2𝐵
𝐵 =
1
2
𝐴𝑙𝑠𝑜, 𝐷 =
1
2
Putting, 𝐷 =
1
2
in (v)
𝐶 + 𝐷 = 0
𝐶 +
1
2
= 0
𝐶 = −
1
2
Putting, 𝐵 =
1
2
in (iv)
1 = 𝐵 + 𝐷

55. 1 =
1
2
+ 𝐷
𝐷 =
1
2
Putting, 𝐶 = −
1
2
in (i)
𝐴 + 𝐶 = 0
𝐴 =
1
2
Now,
𝐼 = ∫
𝑑𝑥
𝑥4+𝑥2+1
𝐼 = ∫ (
𝐴𝑥+𝐵
(𝑥2+𝑥+1)
+
𝐶𝑥+𝐷
(𝑥2−𝑥+1)
) 𝑑𝑥 Using (a)
𝐼 = ∫ (
1
2
𝑥+
1
2
(𝑥2+𝑥+1)
+
−
1
2
𝑥+
1
2
(𝑥2−𝑥+1)
)𝑑𝑥
𝐼 = ∫
1
2
𝑥+
1
2
(𝑥2+𝑥+1)
𝑑𝑥 + ∫
−
1
2
𝑥+
1
2
(𝑥2−𝑥+1)
𝑑𝑥
𝐼 =
1
2
∫
𝑥+1
(𝑥2+𝑥+1)
𝑑𝑥 +
1
2
∫
−𝑥+1
(𝑥2−𝑥+1)
𝑑𝑥
𝐼 =
1
2
∫
𝑥+1
(𝑥2+𝑥+1)
𝑑𝑥 −
1
2
∫
𝑥−1
(𝑥2−𝑥+1)
𝑑𝑥
𝐼 =
1
4
∫
2𝑥+2
(𝑥2+𝑥+1)
𝑑𝑥 −
1
4
∫
2𝑥−2
(𝑥2−𝑥+1)
𝑑𝑥
𝐼 =
1
4
∫
2𝑥+1+1
(𝑥2+𝑥+1)
𝑑𝑥 −
1
4
∫
2𝑥−1−1
(𝑥2−𝑥+1)
𝑑𝑥
𝐼 =
1
4
∫ {
2𝑥+1
(𝑥2+𝑥+1)
+
1
(𝑥2+𝑥+1)
} 𝑑𝑥 −
1
4
∫ {
2𝑥−1
(𝑥2−𝑥+1)
−
1
(𝑥2−𝑥+1)
} 𝑑𝑥
𝐼 =
1
4
∫
2𝑥+1
(𝑥2+𝑥+1)
𝑑𝑥 +
1
4
∫
𝑑𝑥
(𝑥2+𝑥+1)
−
1
4
∫
2𝑥−1
(𝑥2−𝑥+1)
𝑑𝑥 +
1
4
∫
𝑑𝑥
(𝑥2−𝑥+1)
𝐼 =
1
4
∫
2𝑥+1
(𝑥2+𝑥+1)
𝑑𝑥 +
1
4
∫
𝑑𝑥
(𝑥+
1
2
)
2
+(
√3
2
)
2 −
1
4
∫
2𝑥−1
(𝑥2−𝑥+1)
𝑑𝑥 +
1
4
∫
𝑑𝑥
(𝑥−
1
2
)
2
+(
√3
2
)
2
𝐼 =
1
4
log(𝑥2
+ 𝑥 + 1) +
1
4

1
√3
2
tan−1
(
𝑥+
1
2
√3
2
) −
1
4
log(𝑥2
− 𝑥 + 1) +
1
4

1
√3
2
tan−1
(
𝑥−
1
2
√3
2
)
𝐼 =
1
4
log(𝑥2
+ 𝑥 + 1) +
1
4

2
√3
tan−1 (
2𝑥+1
√3
) −
1
4
log( )(𝑥2
− 𝑥 + 1) +
1
4

2
√3
tan−1 (
2𝑥−1
√3
)
𝐼 =
1
4
log(𝑥2
+ 𝑥 + 1) +
1
2√3
tan−1 (
2𝑥+1
√3
) −
1
4
log(𝑥2
− 𝑥 + 1) +
1
2√3
tan−1 (
2𝑥−1
√3
)

56. 𝐼 =
1
4
log(𝑥2
+ 𝑥 + 1) −
1
4
log(𝑥2
− 𝑥 + 1) +
1
2√3
tan−1 (
2𝑥+1
√3
) +
1
2√3
tan−1 (
2𝑥−1
√3
)
𝐼 =
1
4
log
𝑥2+𝑥+1
𝑥2−𝑥+1
+
1
2√3
(tan−1 (
2𝑥+1
√3
) + tan−1 (
2𝑥−1
√3
) )
𝐼 =
1
4
log
𝑥2+𝑥+1
𝑥2−𝑥+1
+
1
2√3
tan−1
(
2𝑥+1
√3
+
2𝑥−1
√3
1−(
2𝑥+1
√3
)(
2𝑥−1
√3
)
)
𝐼 =
1
4
log
𝑥2+𝑥+1
𝑥2−𝑥+1
+
1
2√3
tan−1 (
2𝑥+1+2𝑥−1
√3
1−(
4𝑥2−1
3
)
)
𝐼 =
1
4
log
𝑥2+𝑥+1
𝑥2−𝑥+1
+
1
2√3
tan−1
(
4𝑥
√3
3−4𝑥2+1
3
)
𝐼 =
1
4
log
𝑥2+𝑥+1
𝑥2−𝑥+1
+
1
2√3
tan−1
(
4𝑥
√3
4−4𝑥2
3
)
𝐼 =
1
4
log
𝑥2+𝑥+1
𝑥2−𝑥+1
+
1
2√3
tan−1 (
4𝑥
√3

3
4−4𝑥2
)
𝐼 =
1
4
log
𝑥2+𝑥+1
𝑥2−𝑥+1
+
1
2√3
tan−1 (
4√3 𝑥
4(1−𝑥2)
)
𝐼 =
1
4
log
𝑥2+𝑥+1
𝑥2−𝑥+1
+
1
2√3
tan−1 (
√3 𝑥
(1−𝑥2)
)
27. (i) ∫
𝑑𝑥
𝑥2+1
(ii) ∫
𝑥2+1
𝑥4+1
𝑑𝑥
(i) Solution: Let 𝐼 = ∫
𝑑𝑥
𝑥2+1
𝐼 = ∫
𝑑𝑥
(𝑥2+𝑥√2+1)(𝑥2−𝑥√2+1)
Let,
1
(𝑥2+𝑥√2+1)(𝑥2−𝑥√2+1)
=
𝐴𝑥+𝐵
(𝑥2+𝑥√2+1)
+
𝐶𝑥+𝐷
(𝑥2−𝑥√2+1)
----- (a)
1
(𝑥2+𝑥√2+1)(𝑥2−𝑥√2+1)
=
(𝐴𝑥+𝐵)(𝑥2−𝑥√2+1)+(𝐶𝑥+𝐷)(𝑥2+𝑥√2+1)
(𝑥2+𝑥√2+1)(𝑥2−𝑥√2+1)
1 = (𝐴𝑥 + 𝐵)(𝑥2
− 𝑥√2 + 1) + (𝐶𝑥 + 𝐷)(𝑥2
+ 𝑥√2 + 1)
1 = 𝐴(𝑥3
− 𝑥2
√2 + 𝑥) + 𝐵(𝑥2
− 𝑥√2 + 1) + 𝐶(𝑥3
+ 𝑥2
√2 + 𝑥) + 𝐷(𝑥2
+ 𝑥√2 + 1) ----- (b)
Equating the coefficients of 𝑥3
on both sides of (b), we get
𝐴 + 𝐶 = 0 ------ (i)
Equating the coefficients of 𝑥2
on both sides of (b), we get
𝐵 + 𝐷 + √2 (– 𝐴 + 𝐶) = 0 ------- (ii)
tan−1
𝐴 + tan−1
𝐵 = tan−1
𝐴 + 𝐵
1 − 𝐴𝐵

57. Equating the coefficients of 𝑥 on both sides of (b), we get
𝐵 − 𝐷 = 0 ---- (iii)
Equating the constant terms on both sides of (b), we get
𝐵 + 𝐷 = 1 ----- (iv)
Solving (iii) and (iv), we get
𝐵 =
1
2
Putting, 𝐵 =
1
2
in (iii)
𝐵 − 𝐷 = 0
𝐷 = 𝐵
𝐷 =
1
2
Putting 𝐵 + 𝐷 = 1 in (ii)
𝐵 + 𝐷 + √2 (– 𝐴 + 𝐶) = 0
1 + √2 (– 𝐴 + 𝐶) = 0
1 + √2 (−𝐴 − 𝐴) = 0 Since 𝐴 + 𝐶 = 0 𝐶 = −𝐴
1 − 2√2 𝐴 = 0
𝐴 =
1
2√2
Again, from (i), we have
𝐴 + 𝐶 = 0
𝐶 +
1
2√2
= 0
𝐶 = −
1
2√2
Now,
𝐼 = ∫
𝑑𝑥
(𝑥2+𝑥√2+1)(𝑥2−𝑥√2+1)
𝐼 = ∫ (
𝐴𝑥+𝐵
(𝑥2+𝑥√2+1)
+
𝐶𝑥+𝐷
(𝑥2−𝑥√2+1)
) 𝑑𝑥 Using (a)
𝐼 = ∫
𝐴𝑥+𝐵
(𝑥2+𝑥√2+1)
𝑑𝑥 + ∫
𝐶𝑥+𝐷
(𝑥2−𝑥√2+1)
𝑑𝑥
𝐼 = ∫
1
2√2
𝑥+
1
2
(𝑥2+𝑥√2+1)
𝑑𝑥 + ∫
−
1
2√2
𝑥+
1
2
(𝑥2−𝑥√2+1)
𝑑𝑥
𝐼 =
1
2√2
∫
𝑥+√2
(𝑥2+𝑥√2+1)
𝑑𝑥 −
1
2√2
∫
𝑥−√2
(𝑥2−𝑥√2+1)
𝑑𝑥
𝐼 =
1
4√2
∫
2𝑥+2√2
(𝑥2+𝑥√2+1)
𝑑𝑥 −
1
4√2
∫
2𝑥−2√2
(𝑥2−𝑥√2+1)
𝑑𝑥
𝐼 = −
1
4√2
∫
2𝑥+√2+√2
(𝑥2+𝑥√2+1)
𝑑𝑥 −
1
4√2
∫
2𝑥−√2−√2
(𝑥2−𝑥√2+1)
𝑑𝑥
𝐼 = −
1
4√2
∫
𝑥+√2
(𝑥2+𝑥√2+1)
𝑑𝑥 +
1
4√2
∫
√2
(𝑥2+𝑥√2+1)
𝑑𝑥 −
1
4√2
∫
𝑥−√2
(𝑥2−𝑥√2+1)
𝑑𝑥 +
1
4√2
∫
√2
(𝑥2−𝑥√2+1)
𝑑𝑥
𝐼 = −
1
4√2
∫
𝑥+√2
(𝑥2+𝑥√2+1)
𝑑𝑥 +
1
4
∫
𝑑𝑥
(𝑥2+𝑥√2+1)
−
1
4√2
∫
𝑥−√2
(𝑥2−𝑥√2+1)
𝑑𝑥 +
1
4
∫
𝑑𝑥
(𝑥2−𝑥√2+1)

58. 𝐼 =
1
4√2
log(𝑥2
+ 𝑥√2 + 1) +
1
4
∫
𝑑𝑥
(𝑥+
√2
2
)
2
+(
1
√2
)
2
−
1
4√2
log(𝑥2
− 𝑥√2 + 1) +
1
4
∫
𝑑𝑥
(𝑥−
√2
2
)
2
+(
1
√2
)
2
𝐼 =
1
4√2
{log(𝑥2
+ 𝑥√2 + 1) − log(𝑥2
− 𝑥√2 + 1)} +
1
4

1
1
√2
tan−1
(
𝑥+
√2
2
1
√2
) +
1
4

1
1
√2
tan−1
(
𝑥−
√2
2
1
√2
)
𝐼 =
1
4√2
log
𝑥2+𝑥√2+1
𝑥2−𝑥√2+1
+
√2
4
tan−1 (
2𝑥+√2
√2
) +
√2
4
tan−1 (
2𝑥−√2
√2
)
𝐼 =
1
4√2
log
𝑥2+𝑥√2+1
𝑥2−𝑥√2+1
+
1
2√2
tan−1 (
2𝑥+√2
√2
) +
1
2√2
tan−1 (
2𝑥−√2
√2
)
𝐼 =
1
4√2
log
𝑥2+𝑥√2+1
𝑥2−𝑥√2+1
+
1
2√2
(tan−1 (
2𝑥+√2
√2
) + tan−1 (
2𝑥−√2
√2
))
𝐼 =
1
4√2
log
𝑥2+𝑥√2+1
𝑥2−𝑥√2+1
+
1
2√2
(tan−1 (
2𝑥+√2
√2
+
2𝑥−√2
√2
1−(
2𝑥+√2
√2
)(
2𝑥−√2
√2
)
))
𝐼 =
1
4√2
log
𝑥2+𝑥√2+1
𝑥2−𝑥√2+1
+
1
2√2
(tan−1 (
2𝑥+√2 +2𝑥−√2
√2
1−(
4𝑥2−2
2
)
))
𝐼 =
1
4√2
log
𝑥2+𝑥√2+1
𝑥2−𝑥√2+1
+
1
2√2
(tan−1 (
2𝑥+√2 +2𝑥−√2
√2
2−4𝑥2+2
2
))
𝐼 =
1
4√2
log
𝑥2+𝑥√2+1
𝑥2−𝑥√2+1
+
1
2√2
(tan−1
(
2𝑥 +2𝑥
√2
4−4𝑥2
2
))
𝐼 =
1
4√2
log
𝑥2+𝑥√2+1
𝑥2−𝑥√2+1
+
1
2√2
(tan−1 (
4𝑥
√2

2
4−4𝑥2
))
𝐼 =
1
4√2
log
𝑥2+𝑥√2+1
𝑥2−𝑥√2+1
+
1
2√2
(tan−1 (
4𝑥
√2

2
4(1−𝑥2)
))
𝐼 =
1
4√2
log
𝑥2+𝑥√2+1
𝑥2−𝑥√2+1
+
1
2√2
(tan−1 (
√2 𝑥
(1−𝑥2)
))
𝐼 =
1
4√2
log
𝑥2+𝑥√2+1
𝑥2−𝑥√2+1
+
1
2√2
tan−1 (
√2 𝑥
1−𝑥2
)
(ii) Solution: Let:, 𝐼 = ∫
𝑥2+1
𝑥4+1
𝑑𝑥
𝐼 = ∫
𝑥2+1
(𝑥2+𝑥√2+1)(𝑥2−𝑥√2+1)
𝑑𝑥
Let,
𝑥2+1
(𝑥2+𝑥√2+1)(𝑥2−𝑥√2+1)
=
𝐴𝑥+𝐵
(𝑥2+𝑥√2+1)
+
𝐶𝑥+𝐷
(𝑥2−𝑥√2+1)
----- (a)
𝑥2+1
(𝑥2+𝑥√2+1)(𝑥2−𝑥√2+1)
=
(𝐴𝑥+𝐵)(𝑥2−𝑥√2+1)+(𝐶𝑥+𝐷)(𝑥2+𝑥√2+1)
(𝑥2+𝑥√2+1)(𝑥2−𝑥√2+1)
tan−1
𝐴 + tan−1
𝐵 = tan−1 𝐴+𝐵
1−𝐴𝐵

59. 𝑥2
+ 1 = (𝐴𝑥 + 𝐵)(𝑥2
− 𝑥√2 + 1) + (𝐶𝑥 + 𝐷)(𝑥2
+ 𝑥√2 + 1)
𝑥2
+ 1 = 𝐴(𝑥3
− 𝑥2
√2 + 𝑥) + 𝐵(𝑥2
− 𝑥√2 + 1) + 𝐶(𝑥3
+ 𝑥2
√2 + 𝑥) + 𝐷(𝑥2
+ 𝑥√2 + 1) ----(b)
Equating the coefficients of 𝑥3
on both sides of (b), we get
𝐴 + 𝐶 = 0 ------ (i)
Equating the coefficients of 𝑥2
on both sides of (b), we get
𝐵 + 𝐷 + √2 (– 𝐴 + 𝐶) = 1 ------- (ii)
Equating the coefficients of 𝑥 on both sides of (b), we get
𝐵 − 𝐷 = 0 ---- (iii)
Equating the constant terms on both sides of (b), we get
𝐵 + 𝐷 = 1 ----- (iv)
Solving (iii) and (iv), we get
𝐵 =
1
2
Putting, 𝐵 =
1
2
in (iii)
𝐵 − 𝐷 = 0
𝐷 = 𝐵
𝐷 =
1
2
Putting 𝐵 + 𝐷 = 1 in (ii)
𝐵 + 𝐷 + √2 (– 𝐴 + 𝐶) = 1
1 + √2 (– 𝐴 + 𝐶) = 1
1 + √2 (−𝐴 − 𝐴) = 1 Since 𝐴 + 𝐶 = 0 𝐶 = −𝐴
1 − 2√2 𝐴 = 1
𝐴 = 0
Again, from (i), we have
𝐴 + 𝐶 = 0
𝐶 + 0 = 0
𝐶 = 0
Now,
𝐼 = ∫
𝑥2+1
(𝑥2+𝑥√2+1)(𝑥2−𝑥√2+1)
𝑑𝑥
𝐼 = ∫ (
𝐴𝑥+𝐵
(𝑥2+𝑥√2+1)
+
𝐶𝑥+𝐷
(𝑥2−𝑥√2+1)
) 𝑑𝑥 Using (a)
𝐼 = ∫ (
𝐵
(𝑥2+𝑥√2+1)
+
𝐷
(𝑥2−𝑥√2+1)
) 𝑑𝑥 Since, 𝐴 = 0, 𝐶 = 0
𝐼 = ∫
𝐵
(𝑥2+𝑥√2+1)
𝑑𝑥 + ∫
𝐷
(𝑥2−𝑥√2+1)
𝑑𝑥
𝐼 = 𝐵 ∫
𝑑𝑥
(𝑥+
√2
2
)
2
+(
1
√2
)
2
+ 𝐷 ∫
𝑑𝑥
(𝑥−
√2
2
)
2
+(
1
√2
)
2

60. 𝐼 = 𝐵
1
1
√2
tan−1
(
𝑥+
√2
2
1
√2
) + 𝐷
1
1
√2
tan−1
(
𝑥−
√2
2
1
√2
)
𝐼 = 𝐵
1
1
√2
tan−1 (
2𝑥+√2
√2
) + 𝐷
1
1
√2
tan−1 (2
𝑥−√2
√2
)
𝐼 =
1
2
√2 tan−1 (
2𝑥+√2
√2
) +
1
2
√2 tan−1 (
2𝑥−√2
√2
)
𝐼 =
1
√2
tan−1 (
2𝑥+√2
√2
) +
1
√2
tan−1 (
2𝑥−√2
√2
)
𝐼 =
1
√2
tan−1 (
2𝑥+√2
√2
+
2𝑥−√2
√2
1−(
2𝑥+√2
√2
)(
2𝑥−√2
√2
)
)
𝐼 =
1
√2
tan−1 (
4𝑥
√2
1−(
4𝑥2−2
2
)
)
𝐼 =
1
√2
tan−1
(
4𝑥
√2
2−4𝑥2+2
2
)
𝐼 =
1
√2
tan−1
(
4𝑥
√2
4−4𝑥2
2
)
𝐼 =
1
√2
tan−1
(
4𝑥
√2
4(1−𝑥2)
2
)
𝐼 =
1
√2
tan−1 (
4𝑥
√2

2
4(1−𝑥2)
)
𝐼 =
1
√2
tan−1 (
√2 𝑥
1−𝑥2
)
28. (i) ∫
𝑑𝑥
cos𝑥 (5+3 cos 𝑥)
(ii) ∫
𝑑𝑥
sin 2𝑥−sin 𝑥
(i) Solution: Let, 𝐼 = ∫
𝑑𝑥
cos𝑥 (5+3 cos 𝑥)
Let,
1
cos𝑥 (5+3 cos 𝑥)
=
𝐴
cos𝑥
+
𝐵
5+3cos 𝑥
1
cos 𝑥 (5+3 cos𝑥)
=
𝐴 (5+3 cos 𝑥)+ 𝐵 cos𝑥
cos𝑥 (5+3 cos 𝑥)
1 = 𝐴 (5 + 3 cos 𝑥) + 𝐵 cos 𝑥
1 = 5𝐴 + 3 𝐴 cos 𝑥 + 𝐵 cos 𝑥 ------- (b)
tan−1
𝐴 + tan−1
𝐵 = tan−1 𝐴+𝐵
1−𝐴𝐵

61. Equating the coefficients of cos 𝑥 on both sides of equation (b), we get
0 = 3𝐴 + 𝐵
 𝐵 = −3𝐴 ------ (i)
Equating the constant terms on both sides of equation (b), w e get
1 = 5𝐴
𝐴 =
1
5
------ (ii)
Putting 𝐴 =
1
5
in equation (i), we get
𝐵 = −3𝐴
𝐵 = −
3
5
Now,
𝐼 = ∫
𝑑𝑥
cos 𝑥 (5+3 cos𝑥)
𝐼 = ∫ (
𝐴
cos 𝑥
+
𝐵
5+3 cos𝑥
) 𝑑𝑥
𝐼 = ∫
𝐴
cos 𝑥
𝑑𝑥 + ∫
𝐵
5+3 cos 𝑥
𝑑𝑥
𝐼 = 𝐴 ∫
1
cos 𝑥
𝑑𝑥 + 𝐵 ∫
1
5+3 cos 𝑥
𝑑𝑥
𝐼 = 𝐴 log |tan (
𝜋
4
+
𝑥
2
)| + 𝐵 ∫
1
5+3(cos2𝑥
2
−sin2𝑥
2
)
𝑑𝑥
𝐼 = 𝐴 log |tan (
𝜋
4
+
𝑥
2
)| + 𝐵 ∫
1
5(sin2𝑥
2
+cos2𝑥
2
)+3 cos2𝑥
2
− 3 sin2𝑥
2
𝑑𝑥
𝐼 = 𝐴 log |tan (
𝜋
4
+
𝑥
2
)| + 𝐵 ∫
1
5 sin2𝑥
2
+ 5cos2𝑥
2
+3 cos2𝑥
2
− 3 sin2𝑥
2
𝑑𝑥
𝐼 = 𝐴 log |tan (
𝜋
4
+
𝑥
2
)| + 𝐵 ∫
1
2 sin2𝑥
2
+ 8cos2𝑥
2
𝑑𝑥
𝐼 = 𝐴 log |tan (
𝜋
4
+
𝑥
2
)| + 𝐵 ∫
1
2 cos2𝑥
2
(
sin2𝑥
2
cos2𝑥
2
+ 4)
𝑑𝑥
𝐼 = 𝐴 log |tan (
𝜋
4
+
𝑥
2
)| + 𝐵 ∫
1
2 cos2𝑥
2
(tan2𝑥
2
+ 4)
𝑑𝑥
𝐼 = 𝐴 log |tan (
𝜋
4
+
𝑥
2
)| +
𝐵
2
∫
sec2𝑥
2
tan2𝑥
2
+ 4
𝑑𝑥
Take,tan
𝑥
2
= 𝑢  sec2 𝑥
2
𝑑𝑥 = 2 𝑑𝑢

62. 𝐼 = 𝐴 log |tan (
𝜋
4
+
𝑥
2
)| +
𝐵
2
∫
2 𝑑𝑢
𝑢2 + 4
𝐼 = 𝐴 log |tan (
𝜋
4
+
𝑥
2
)| + 𝐵 ∫
𝑑𝑢
𝑢2 + 22
𝐼 = 𝐴 log |tan (
𝜋
4
+
𝑥
2
)| + 𝐵
1
2
tan−1 𝑢
2
𝐼 = 𝐴 log |tan (
𝜋
4
+
𝑥
2
)| + 𝐵
1
2
tan−1 𝑢
2
𝐼 = 𝐴 log |tan (
𝜋
4
+
𝑥
2
)| + 𝐵
1
2
tan−1 (
1
2
𝑢)
𝐼 =
1
5
log |tan(
𝜋
4
+
𝑥
2
)| −
3
5

1
2
tan−1 (
1
2
tan
𝑥
2
)
𝐼 =
1
5
log |tan(
𝜋
4
+
𝑥
2
)| −
3
10
tan−1 (
1
2
tan
𝑥
2
)
(ii) Solution : Let, 𝐼 = ∫
𝑑𝑥
sin 2𝑥−sin 𝑥
𝐼 = ∫
𝑑𝑥
2 sin 𝑥 cos 𝑥−sin 𝑥
Multiplying both denominator and numerator by sin 𝑥, we get
𝐼 = ∫
sin 𝑥 𝑑𝑥
2 sin2 𝑥 cos 𝑥− sin2 𝑥
𝐼 = ∫
sin 𝑥 𝑑𝑥
sin2 𝑥(2 cos 𝑥−1)
𝐼 = ∫
sin 𝑥 𝑑𝑥
− sin2 𝑥(1−2 cos 𝑥)
𝐼 = ∫
− sin 𝑥 𝑑𝑥
(1−cos2 𝑥)(1−2cos 𝑥)
Take, 𝑢 = cos 𝑥  − sin 𝑥 𝑑𝑥 = 𝑑𝑢
𝐼 = ∫
𝑑𝑢
(1− 𝑢2)(1−2𝑢)
𝐼 = ∫
𝑑𝑢
(1+𝑢) (1− 𝑢)(1−2𝑢)
Let,
1
(1+𝑢) (1− 𝑢)(1−2𝑢)
=
𝐴
(1+𝑢)
+
𝐵
(1− 𝑢)
+
𝐶
(1−2𝑢)
---- (a)
1
(1+𝑢) (1− 𝑢)(1−2𝑢)
=
𝐴 (1− 𝑢)(1−2𝑢)+ B (1+𝑢) (1−2𝑢)+ 𝐶(1+𝑢) (1− 𝑢)
(1+𝑢) (1− 𝑢)(1−2𝑢)
1 = 𝐴 (1 − 𝑢)(1 − 2𝑢) + B (1 + 𝑢) (1 − 2𝑢) + 𝐶(1 + 𝑢) (1 − 𝑢) ---------- (b)
Putting, 𝑢 = 1 in (b), we get
1 = −2𝐵

63.  𝐵 = −
1
2
Putting, 𝑢 = −1 in (b), we get
1 = 6𝐴
 𝐴 =
1
6
Putting, 𝑢 =
1
2
in (b), we get
1 =
3
4
𝐶
𝐶 =
4
3
Now,
𝐼 = ∫
𝑑𝑢
(1+𝑢) (1− 𝑢)(1−2𝑢)
𝐼 = ∫ (
𝐴
(1+𝑢)
+
𝐵
(1− 𝑢)
+
𝐶
(1−2𝑢)
)𝑑𝑢 Using (a)
𝐼 = ∫
𝐴
(1+𝑢)
𝑑𝑢 + ∫
𝐵
(1− 𝑢)
𝑑𝑢 + ∫
𝐶
(1−2𝑢)
𝐼 = ∫
𝐴
(1+𝑢)
𝑑𝑢 + ∫
𝐵
(1− 𝑢)
𝑑𝑢 + ∫
𝐶
(1−2𝑢)
𝐼 = 𝐴 ∫
𝑑𝑢
(1+𝑢)
+ 𝐵 ∫
𝑑𝑢
(1− 𝑢)
+ 𝐶 ∫
𝑑𝑢
(1−2𝑢)
𝐼 = 𝐴 log(1 + 𝑢) − 𝐵 log(1 − 𝑢) −
𝐶
2
log(1 − 2𝑢)
𝐼 =
1
6
log(1 + cos 𝑥) +
1
2
log(1 − cos 𝑥) −
4
3
2
log(1 − 2cos 𝑥)
𝐼 =
1
6
log(1 + cos 𝑥) +
1
2
log(1 − cos 𝑥) −
4
6
log(1 − 2cos 𝑥)
𝐼 =
1
6
log(1 + cos 𝑥) +
1
2
log(1 − cos 𝑥) −
2
3
log(1 − 2 cos 𝑥)
𝐼 =
1
6
log(1 + cos 𝑥) +
1
2
log(1 − cos 𝑥) −
2
3
log(1 − 2 cos 𝑥)
𝐼 =
1
6
log(1 + cos 𝑥) +
1
2
log(1 − cos 𝑥) −
2
3
log(1 − 2 cos 𝑥) + 𝐶
29. (i) ∫
𝑑𝑥
1+3𝑒𝑥+2𝑒2𝑥 (ii) ∫
𝑒𝑥 𝑑𝑥
𝑒𝑥−3𝑒−𝑥+2
(i) Solution: Let , 𝐼 = ∫
𝑑𝑥
1+3𝑒𝑥+2𝑒2𝑥
𝐼 = ∫
𝑒𝑥 𝑑𝑥
𝑒𝑥(1+3𝑒𝑥+2𝑒2𝑥)

64. 𝐼 = ∫
𝑒𝑥 𝑑𝑥
𝑒𝑥+3𝑒2𝑥+2𝑒3𝑥
Take, 𝑒𝑥
= 𝑢  𝑒𝑥
𝑑𝑥 = 𝑑𝑢
𝐼 = ∫
𝑑𝑢
𝑢+3𝑢2+2𝑢3
𝐼 = ∫
𝑑𝑢
𝑢(1+3𝑢+2𝑢2)
𝐼 = ∫
𝑑𝑢
𝑢 (1+𝑢)(1+2𝑢)
Let,
𝑑𝑢
𝑢 (1+𝑢)(1+2𝑢)
=
𝐴
𝑢
+
𝐵
(1+𝑢)
+
𝐶
(1+2𝑢)
------ (a)
𝑑𝑢
𝑢 (1+𝑢)(1+2𝑢)
=
𝐴 (1+𝑢)(1+2𝑢)+𝐵 𝑢 (1+2𝑢)+ 𝐶𝑢(1+𝑢)
𝑢 (1+𝑢)(1+2𝑢)
1 = 𝐴 (1 + 𝑢)(1 + 2𝑢) + 𝐵 𝑢 (1 + 2𝑢) + 𝐶 𝑢(1 + 𝑢) ------ (b)
Putting 𝑢 = 0 in (b), we get
𝐴 = 1 ----- (i)
Putting 𝑢 = −1 in (b), we get
𝐵 = 1 ----- (ii)
Putting 𝑢 = −
1
2
in (b), we get
𝐶 = −4
Now,
𝐼 = ∫
𝑑𝑢
𝑢 (1+𝑢)(1+2𝑢)
𝐼 = ∫ (
𝐴
𝑢
+
𝐵
(1+𝑢)
+
𝐶
(1+2𝑢)
) 𝑑𝑢
𝐼 = ∫
𝐴
𝑢
𝑑𝑢 + ∫
𝐵
(1+𝑢)
𝑑𝑢 + ∫
𝐶
(1+2𝑢)
𝑑𝑢
𝐼 = 𝐴 ∫
𝑑𝑢
𝑢
+ 𝐵 ∫
𝑑𝑢
(1+𝑢)
+ 𝐶 ∫
𝑑𝑢
(1+2𝑢)
𝐼 = 𝐴 log 𝑢 + 𝐵 log(1 + 𝑢) +
1
2
𝐶 log(1 + 2𝑢)
𝐼 = log 𝑒𝑥
+ log(1 + 𝑒𝑥) +
−4
2
log(1 + 2𝑒𝑥)
𝐼 = 𝑥 log 𝑒 + log(1 + 𝑒𝑥) − 2 log(1 + 2𝑒𝑥)
𝐼 = 𝑥 + log(1 + 𝑒𝑥) − 2 log(1 + 2𝑒𝑥)

65. (ii) Let, 𝐼 = ∫
𝑒𝑥𝑑𝑥
𝑒𝑥−3𝑒−𝑥+2
𝐼 = ∫
𝑒𝑥𝑒𝑥𝑑𝑥
𝑒𝑥 (𝑒𝑥−3𝑒−𝑥+2)
𝐼 = ∫
𝑒𝑥𝑒𝑥𝑑𝑥
𝑒2𝑥+2𝑒𝑥 −3
Take, 𝑒𝑥
= 𝑢  𝑒𝑥
𝑑𝑥 = 𝑑𝑢
𝐼 = ∫
𝑢 𝑑𝑢
𝑢2+2𝑢 −3
𝐼 = ∫
𝑢 𝑑𝑢
𝑢2+3𝑢−𝑢 −3
𝐼 = ∫
𝑢 𝑑𝑢
𝑢(𝑢+3)−1(𝑢+3)
𝐼 = ∫
𝑢 𝑑𝑢
(𝑢−1)(𝑢+3)
Let,
𝑢
(𝑢−1)(𝑢+3)
=
𝐴
(𝑢−1)
+
𝐵
(𝑢+3)
----- (a)
𝑢
(𝑢−1)(𝑢+3)
=
𝐴(𝑢+3)+ 𝐵 (𝑢−1)
(𝑢−1)(𝑢+3)
𝑢 = 𝐴(𝑢 + 3) + 𝐵 (𝑢 − 1) ----- (b)
Putting, 𝑢 = −3 in (b), we get
𝐵 =
3
4
Putting, 𝑢 = 1 in (b), we get
𝐴 =
1
4
Now,
𝐼 = ∫
𝑑𝑢
(𝑢−1)(𝑢+3)
𝐼 = ∫ (
𝐴
(𝑢−1)
+
𝐵
(𝑢+3)
) 𝑑𝑢
𝐼 = ∫
𝐴
(𝑢−1)
𝑑𝑢 + ∫
𝐵
(𝑢+3)
𝑑𝑢
𝐼 = 𝐴 ∫
𝑑𝑢
(𝑢−1)
+ 𝐵 ∫
𝑑𝑢
(𝑢+3)
𝐼 = 𝐴 log(𝑢 − 1) + 𝐵 log(𝑢 + 3)
𝐼 =
1
4
log(𝑒𝑥
− 1) +
3
4
log(𝑒𝑥
+ 3)
𝐼 =
1
4
log(𝑒𝑥
− 1) +
1
4
log(𝑒𝑥
+ 3)3

66. 𝐼 =
1
4
(log(𝑒𝑥
− 1) + log(𝑒𝑥
+ 3)3)
𝐼 =
1
4
log{(𝑒𝑥
− 1)(𝑒𝑥
+ 3)3}
30. ∫
𝑑𝑥
sin 𝑥 (3+2 cos𝑥)
Solution: Let, 𝐼 = ∫
𝑑𝑥
sin 𝑥 (3+2 cos𝑥)
𝐼 = ∫
sin 𝑥 𝑑𝑥
sin2 𝑥 (3+2 cos 𝑥)
On multiplying both numerator and denominator by sin 𝑥
𝐼 = ∫
sin 𝑥 𝑑𝑥
(1−cos2 𝑥) (3+2 cos 𝑥)
Taking cos 𝑥 = 𝑢  sin𝑥 𝑑𝑥 = −𝑑𝑢
𝐼 = ∫
−𝑑𝑢
(1−u2) (3+2𝑢)
𝐼 = ∫
−𝑑𝑢
−(u2−1) (3+2𝑢)
𝐼 = ∫
−𝑑𝑢
(u2−1) (3+2𝑢)
𝐼 = ∫
𝑑𝑢
(1+𝑢)(1−u) (3+2𝑢)
Let,
−1
(𝑢+1)(u−1) (3+2𝑢)
=
𝐴
(1+𝑢)
+
𝐵
(1−𝑢)
+
𝐶
(3+2𝑢)
------- (a)
1
(1+𝑢)(1−u) (3+2𝑢)
=
𝐴(3+2𝑢)(1−𝑢)+ 𝐵(1+𝑢)(3+2𝑢)+ 𝐶 (1+𝑢)(1−𝑢)
(1−𝑢)(1+𝑢)(3+2𝑢)
−1 = 𝐴(3 + 2𝑢)(1 − 𝑢) + 𝐵(1 + 𝑢)(3 + 2𝑢) + 𝐶 (1 + 𝑢)(1 − 𝑢) -------- (b)
Putting, 𝑢 = 1 in (b), we get
1 = 10𝐵
 𝐵 = −
1
10
Putting, 𝑢 = −1 in (b), we get
𝐴 = −
1
2
Putting, 𝑢 = −
3
2
in (b), we get
𝐶 =
4
5
Now,

67. 𝐼 = ∫
−𝑑𝑢
(1+𝑢)(1−u) (3+2𝑢)
𝐼 = ∫ (
𝐴
(1+𝑢)
+
𝐵
(1−𝑢)
+
𝐶
(3+2𝑢)
) 𝑑𝑢
𝐼 = ∫
𝐴
(1+𝑢)
𝑑𝑢 + ∫
𝐵
(1−𝑢)
𝑑𝑢 + ∫
𝐶
(3+2𝑢)
𝑑𝑢
𝐼 = 𝐴 ∫
𝑑𝑢
(1+𝑢)
+ 𝐵 ∫
𝑑𝑢
(1−𝑢)
+ 𝐶 ∫
𝑑𝑢
(3+2𝑢)
𝐼 = 𝐴 log(1 + 𝑢) − 𝐵 log(1 − 𝑢) +
1
2
𝐶 log(3 + 2𝑢)
𝐼 = −
1
2
log(cos 𝑥 + 1) +
1
10
log(1 − cos 𝑥) +
1
2

4
5
log(3 + 2 cos 𝑥)
𝐼 = −
1
2
log(1 + cos 𝑥) +
1
10
log(1 − cos 𝑥) +
2
5
log(3 + 2 cos 𝑥)