The document summarizes techniques for evaluating 10 different types of integrals: 1) Inverse trigonometric substitution 2) Logarithmic substitution 3) Algebraic substitution 4) Trigonometric substitution 5) Exponential substitution 6) Partial fraction decomposition 7) Trigonometric identities 8) Trigonometric angle addition identities 9) Rationalizing the denominator 10) Partial fraction decomposition of a rational function

1. Republica Bolivariana de Venezuela
Ministerio del Poder Popular para la Educación
Instituto Universitario Politécnico “ Santiago Mariño”
Ingeniería Civil
Vivas Jornely
C.I 27.239.169
Matemática II
Sección c

2. (1)
𝑥2
𝑥3+8
𝑑𝑥 u=𝑋3
+ 8 du = 3𝑋2
dx
𝑑𝑢
3
= 𝑥2
𝑑𝑥
1
𝑢
×
𝑑𝑢
3
=
𝑑𝑢
𝑢 × 3
=
1
3
×
𝑑𝑢
𝑢
=
1
3
𝑑𝑢
𝑢
=
1
3
ln 𝑥3 + 8 + 𝑐
(2)
3𝑥 𝑑𝑥
3
𝑥2+2
𝑢 = 𝑥2
+ 2; 𝑑𝑢 = 2𝑥 𝑑𝑥 ;
𝑑𝑢
3
= 𝑥3
𝑑𝑥
𝑑𝑢
𝑢
= ln 𝑢 + 𝑐
3𝑥 𝑑𝑥
𝑥2+2 1
3
= 3𝑥 × 𝑥2+2 −1
3 𝑑𝑥 = 𝑢
−1
3 × 3𝑥 𝑑𝑥 =
𝑢
−1
3 × 3𝑥
𝑑𝑢
2𝑥
= 𝑢−1 3 ×
3
2
𝑑𝑢 =
3
2
𝑢
−1
3 𝑑𝑢 =
3
2
𝑢
2
3
2
3
+ 𝑐 =
3
2
3𝑢
2
3
2
+ 𝑐 =
9
4
𝑢
2
3 + 𝑐 =
9
3
𝑥2 + 2 2
4
+ c
𝑢2 𝑑𝑢 =
𝑢 𝑛+1
𝑛 + 1
+ 𝑐

3. I nversa
L ogaritmica
A lgebraica
T rigonometrica
E xponencial
𝑥𝑒3𝑥 𝑑𝑥 = 𝑥
1
3
𝑒3𝑥 −
1
3
𝑒3𝑥 𝑑𝑥
=
1
3
𝑥𝑒3𝑥
−
1
3
𝑒3𝑥
𝑑𝑥
=
1
3
𝑥𝑒3𝑥
−
1
3
1
3
𝑒3𝑥
𝑐
=
1
3
𝑥𝑒3𝑥 −
1
9
𝑒3𝑥 + 𝑐
(3) 𝑥𝑒3𝑥 𝑑𝑥
• 𝑢 = 𝑥
• 𝑑𝑣 = 𝑒3𝑥
𝑑𝑥 = 𝑒3𝑥
𝑑𝑥
𝑣 =
1
3
𝑒3𝑥
• 𝑑𝑢 = 𝑥𝑑𝑥
𝑑𝑢 = 𝑑𝑥
(4) 𝑥𝑡𝑎𝑛−1
𝑥 𝑑𝑥
• 𝑢 = 𝑡𝑎𝑛−1 𝑥
• 𝑑𝑣 = 𝑥𝑑𝑥 = 𝑥𝑑𝑥
𝑣 =
𝑥2
2
• 𝑑𝑢 =
1
1+𝑥2 𝑑𝑥
𝑡𝑎𝑛−1 𝑥 𝑥 𝑑𝑥 = 𝑡𝑎𝑛−1 𝑥
𝑥2
2
−
𝑥2
2
×
1
1 + 𝑥2
𝑑𝑥
=
𝑥2
𝑡𝑎𝑛−1
𝑥
2
−
1
2
𝑥2
1 + 𝑥2
𝑑𝑥
=
𝑥2 𝑡𝑎𝑛−1 𝑥
𝑥
−
1
2
𝑥 − 𝑡𝑎𝑛−1
𝑥 + 𝑐
𝑥2
1 + 𝑥2
𝑑𝑥 =
𝑥2
+ 1 − 1
1 + 𝑥2
𝑑𝑥 =
𝑥2
+ 1
1 + 𝑥2
−
1
1 + 𝑥2
𝑑𝑥 = 1𝑑𝑥 −
1
1 + 𝑥2
𝑑𝑥 = 𝑥 − 𝑡𝑎𝑛−1
𝑥 + 𝑐

4. 𝑐𝑜𝑠𝑥 𝑑𝑐 = 𝑠𝑒𝑛𝑥 + 𝑐
(5)
𝑑𝑥
𝑥2 + 1
3
2
1
x
cos 𝛼 =
1
𝑥2 + 1
𝑥2 + 1 =
1
cos 𝛼
𝑥2 + 1 = sec 𝛼
𝑥2
+ 1
1
2 = sec 𝛼
𝑥2
+ 1
1
2
3
= sec 𝛼 3
𝑥2
+ 1
3
2 = sec 𝛼3
tan 𝛼 =
𝑥
1
tan 𝛼 = 𝑥
𝑑𝑥
𝑑𝛼
= sec 𝛼2
𝑑𝑥 = 𝑠𝑒𝑐2
𝛼 𝑑𝛼
𝑑𝑥
𝑥2 + 1
3
2
=
𝑠𝑒𝑐2 𝛼
𝑠𝑒𝑐3 𝛼
𝑑𝑥 =
1
𝑠𝑒𝑐𝛼
𝑑𝛼 = 𝑐𝑜𝑠𝛼 𝑑𝛼 = 𝑠𝑒𝑛𝛼 + 𝑐 =
𝑥
𝑥2 + 1
+ 𝑐

5. 𝑐𝑜𝑠 𝑘𝛼 𝑑𝛼 =
1
𝑘
𝑠𝑒𝑛 𝑘𝛼
𝑐𝑜𝑠2 𝛼 =
1 + 𝑐𝑜𝑠2𝛼
2
(6) 4 − 𝑥2 𝑑𝑥
2
4 − 𝑥2
x
• cos 𝛼 =
4−𝑥2
2
2cos 𝛼 = 4 − 𝑥2
• sin 𝛼 =
𝑥
2
2 sin 𝛼 = 𝑥
𝑥 = 2 sin 𝛼
𝑑𝑥 = 2𝑐𝑜𝑠𝛼 𝑑𝛼
2𝑐𝑜𝑠𝛼 × 2𝑐𝑜𝑠𝛼 𝑑𝛼 =
4𝑐𝑜𝑠2 𝛼 𝑑𝛼 =
4 𝑐𝑜𝑠2
𝛼 𝑑𝛼 =
4
1 + 𝑐𝑜𝑠2𝛼
2
𝑑𝛼 =
2 1 + 𝑐𝑜𝑠2𝛼 𝑑𝛼 =
2 1 𝑑𝛼 + 𝑐𝑜𝑠2𝛼 𝑑𝛼 =
2 𝛼 +
1
2
𝑠𝑒𝑛 2𝛼 + 𝑐

6. cos 2𝑥 = 1 − 2𝑠𝑒𝑛2
𝑥
cos 𝑘𝑥 𝑑𝑥 =
1
𝑘
𝑠𝑒𝑛 𝑘𝑥 + 𝑐
(7) 𝑠𝑒𝑛2 𝑥𝑑𝑥 =
1−𝑐𝑜𝑠2𝑥
2
𝑑𝑥
=
1
2
1 − 𝑐𝑜𝑠2𝑥 𝑑𝑥
=
1
2
𝑑𝑥 − 𝑐𝑜𝑠2𝑥 𝑑𝑥
=
1
2
𝑥 −
1
2
𝑠𝑒𝑛2𝑥 + 𝑐
=
𝑥
2
−
1
4
𝑠𝑒𝑛2𝑥 + 𝑐
=
𝑥
2
−
𝑠𝑒𝑛2𝑥
4
+ 𝑐
(8)
𝑐𝑜𝑠𝑥
𝑠𝑒𝑛𝑥
𝑑𝑥 =
1
𝑠𝑒𝑛𝑥
×
𝑐𝑜𝑠𝑥
𝑠𝑒𝑛𝑥
𝑑𝑥
= 𝑐𝑠𝑐𝑥 × 𝑐𝑜𝑡𝑥 𝑑𝑥
= −𝑐𝑠𝑐 + 𝑐

7. (9)
6𝑥+13
𝑥+3 𝑥+2
=
𝐴 𝑋+2 +𝐵 𝑋+8
𝑋+3 𝑋+2
6𝑋 + 13 = 𝐴 𝑋 + 2 + 𝐵(𝑋 + 3)
6𝑋 + 13 = 𝐴𝑋 + 2𝐴 + 𝐵𝑋 + 3𝐵
6𝑋 = 𝐴𝑋 + 𝐵𝑋
13 = 2𝐴 + 3𝐵
6 = 𝐴 + 𝐵 −2
13 = 2𝐴 + 3𝐵
−12 = −2𝐴 − 2𝐵
13 = 2𝐴 + 3𝐵
1 = 0 + 𝐵
6 = 𝐴 + 𝐵
6 = 𝐴 + 1
6 − 1 = 𝐴
𝐴 = 5
6𝑋 + 13
𝑋 + 3 𝑋 + 2
𝑑𝑥 =
5
𝑥 + 3
+
1
𝑥 + 2

8. (10)
2𝑥+1
(𝑥+1)(𝑥2+4)
𝑑𝑥 =
𝐴
(𝑥+1)
+
𝐵𝑋+𝐶
(𝑋2+4)
2𝑋+1
(𝑋+1)(𝑋2+4)
=
𝐴 𝑋2+4 +𝐵𝑋+𝐶(𝑋+1)
(𝑋+1)(𝑋2+4)
2𝑋 + 1 = 𝐴 𝑋2
+ 4 + (𝐵𝑋 + 𝐶)(𝑋 +
1)
• X= -1 → −2 + 1 = 𝐴 −1 + 4 + (𝐵 −1 + 𝐶)( −1 + 1)
−1 = 5𝐴 + 𝐶 − 𝐵
−
1
5
= 𝐴
• X=0 → 1 = 4𝐴 + 𝐶
1 = 4 −
1
5
+ 𝐶
1 +
4
5
= 𝐶
9
5
= 𝐶
• X=1 → 3 = 5𝐴 + (𝐵 + 𝐶)(`1 + 1)
3 = 5𝐴 + 2𝐵 + 2𝐶
3 = 5 −
1
5
+ 2𝐵 + 2(
9
5
)
3 = −1 + 2𝐵 +
18
5
4 = 2𝐵 +
18
5
4 −
18
5
= 2𝐵
2
5
= 2𝐵
1
5
= 𝐵

9. 2𝑥 + 1
(𝑥 + 1)(𝑥2 + 4)
𝑑𝑥 =
𝐴
(𝑋 + 1)
+
𝐵𝑋 + 𝐶
(𝑋2 + 4)
=
−1
5
(𝑋+1)
+
1
5 𝑋+9
5
(𝑋2+4)
𝑑𝑥
= −
1
5 𝑋+1
+
1
5
(
𝑋+9
5
(𝑋2+4)
) 𝑑𝑥
=−
1
5
𝑑𝑥
(𝑥+1)
+
1
5
𝑥 𝑑𝑥
(𝑥2+4)
+
1
5
9 𝑑𝑥
𝑥2+4
=−
1
5
ln 𝑥 + 1 +
1
5
𝑙𝑛𝑥2 + 4 +
9
10
𝑎𝑟𝑡𝑔
𝑥
2
+ 𝑐