The document describes the process of integration by partial fractions. It explains that when the degree of the numerator is greater than or equal to the denominator, division is performed. Otherwise, the denominator is factored. For each linear factor, the numerator is written as a sum of terms divided by that factor. For multiple linear factors, the numerator is written as a sum of terms divided by powers of that factor. Examples are provided to demonstrate these steps.

1. Integration By
Partial Fractions
() Ax
To find; ∫ dx
P( x )

2. Integration By
Partial Fractions
() Ax
To find; ∫ dx
P( x )
(1) If degA( x ) ≥ degP( x ) , perform a division

3. Integration By
Partial Fractions
()Ax
To find; ∫ dx
P( x )
(1) If degA( x ) ≥ degP( x ) , perform a division
( 2) If degA( x ) < degP( x ) , factorise P( x )

4. Integration By
Partial Fractions
()Ax
To find; ∫ dx
P( x )
(1) If degA( x ) ≥ degP( x ) , perform a division
( 2) If degA( x ) < degP( x ) , factorise P( x )
A
a) for linear factor ( x − a ) , write
x−a

5. Integration By
Partial Fractions
()Ax
To find; ∫ dx
P( x )
(1) If degA( x ) ≥ degP( x ) , perform a division
( 2) If degA( x ) < degP( x ) , factorise P( x )
A
a) for linear factor ( x − a ) , write
x−a
b) for multiple linear factors ( x − a ) , write
n
A B C
+ ++
( x − a) ( x − a) ( x − a) n
2

6. Integration By
Partial Fractions
()Ax
To find; ∫ dx
P( x )
(1) If degA( x ) ≥ degP( x ) , perform a division
( 2) If degA( x ) < degP( x ) , factorise P( x )
A
a) for linear factor ( x − a ) , write
x−a
b) for multiple linear factors ( x − a ) , write
n
A B C
+ ++
( x − a) ( x − a) ( x − a) n
2
Ax + B
c) for polynomial factors e.g. ax + bx + c, write 2
2
ax + bx + c

7. x2
e.g. ( i ) ∫ dx
x +1

8. x
x2
e.g. ( i ) ∫ dx x +1 x 2 + 0 x + 0
x +1
x2 + x

9. x −1
x2
e.g. ( i ) ∫ dx x +1 x 2 + 0 x + 0
x +1
x2 + x
−x+0

10. x −1
x2
e.g. ( i ) ∫ dx x +1 x 2 + 0 x + 0
x +1
x2 + x
−x+0
− x −1

11. x −1
x2
e.g. ( i ) ∫ dx x +1 x 2 + 0 x + 0
x +1
x2 + x
−x+0
− x −1
1

12. x −1
x2
e.g. ( i ) ∫ dx x +1 x 2 + 0 x + 0
x +1
x2 + x
= ∫ x −1 +
1
 dx
 −x+0
x + 1

− x −1
1

13. x −1
x2
e.g. ( i ) ∫ dx x +1 x 2 + 0 x + 0
x +1
x2 + x
= ∫ x −1 +
1
 dx
 −x+0
x + 1

− x −1
12
= x − x + log( x + 1) + c
1
2

14. x −1
x2
e.g. ( i ) ∫ dx x +1 x 2 + 0 x + 0
x +1
x2 + x
= ∫ x −1 +
1
 dx
 −x+0
x + 1

− x −1
12
= x − x + log( x + 1) + c
1
2
3dx
( ii ) ∫ 2
x −x

15. x −1
x2
e.g. ( i ) ∫ dx x +1 x 2 + 0 x + 0
x +1
x2 + x
= ∫ x −1 +
1
 dx
 −x+0
x + 1

− x −1
12
= x − x + log( x + 1) + c
1
2
3dx
( ii ) ∫ 2
x −x
3dx
=∫
x( x − 1)

16. x −1
x2
e.g. ( i ) ∫ dx x +1 x 2 + 0 x + 0
x +1
x2 + x
= ∫ x −1 +
1
 dx
 −x+0
x + 1

− x −1
12
= x − x + log( x + 1) + c
1
2
A B 3
3dx
( ii ) ∫ 2 + =
x x − 1 x( x − 1)
x −x
3dx
=∫
x( x − 1)

17. x −1
x2
e.g. ( i ) ∫ dx x +1 x 2 + 0 x + 0
x +1
x2 + x
= ∫ x −1 +
1
 dx
 −x+0
x + 1

− x −1
12
= x − x + log( x + 1) + c
1
2
3dx A B 3
( ii ) ∫ 2 + =
x x − 1 x( x − 1)
x −x
3dx A( x − 1) + Bx = 3
=∫
x( x − 1)

18. x −1
x2
e.g. ( i ) ∫ dx x +1 x 2 + 0 x + 0
x +1
x2 + x
= ∫ x −1 +
1
 dx
 −x+0
x + 1

− x −1
12
= x − x + log( x + 1) + c
1
2
3dx A B 3
( ii ) ∫ 2 + =
x x − 1 x( x − 1)
x −x
3dx A( x − 1) + Bx = 3
=∫
x( x − 1)
x=0
−A=3
A = −3

19. x −1
x2
e.g. ( i ) ∫ dx x +1 x 2 + 0 x + 0
x +1
x2 + x
= ∫ x −1 +
1
 dx
 −x+0
x + 1

− x −1
12
= x − x + log( x + 1) + c
1
2
3dx A B 3
( ii ) ∫ 2 + =
x x − 1 x( x − 1)
x −x
3dx A( x − 1) + Bx = 3
=∫
x( x − 1)
x=0 x =1
−A=3 B=3
A = −3

20. x −1
x2
e.g. ( i ) ∫ dx x +1 x 2 + 0 x + 0
x +1
x2 + x
= ∫ x −1 +
1
 dx
 −x+0
x + 1

− x −1
12
= x − x + log( x + 1) + c
1
2
3dx A B 3
( ii ) ∫ 2 + =
x x − 1 x( x − 1)
x −x
3dx A( x − 1) + Bx = 3
=∫
x( x − 1)
x=0 x =1
− 3 3
= ∫ +  dx −A=3 B=3
 x ( x − 1) 
A = −3

21. x −1
x2
e.g. ( i ) ∫ dx x +1 x 2 + 0 x + 0
x +1
x2 + x
= ∫ x −1 +
1
 dx
 −x+0
x + 1

− x −1
12
= x − x + log( x + 1) + c
1
2
3dx A B 3
( ii ) ∫ 2 + =
x x − 1 x( x − 1)
x −x
3dx A( x − 1) + Bx = 3
=∫
x( x − 1)
x=0 x =1
− 3 3
= ∫ +  dx −A=3 B=3
 x ( x − 1) 
A = −3
= −3 log x + 3 log( x − 1) + c

22. x −1
x2
e.g. ( i ) ∫ dx x +1 x 2 + 0 x + 0
x +1
x2 + x
= ∫ x −1 +
1
 dx
 −x+0
x + 1

− x −1
12
= x − x + log( x + 1) + c
1
2
3dx A B 3
( ii ) ∫ 2 + =
x x − 1 x( x − 1)
x −x
3dx A( x − 1) + Bx = 3
=∫
x( x − 1)
x=0 x =1
− 3 3
= ∫ +  dx −A=3 B=3
 x ( x − 1) 
A = −3
= −3 log x + 3 log( x − 1) + c
x − 1
= 3 log +c

x

23. x+5
( iii ) ∫ 2 dx
x − 3 x − 10

24. x+5
( iii ) ∫ 2 dx
x − 3 x − 10
x+5
=∫ dx
( x − 5)( x + 2)

25. x+5 x+5
A B
( iii ) ∫ 2 dx + =
( x − 5) ( x + 2) ( x − 5)( x + 2)
x − 3 x − 10
x+5
A( x + 2 ) + B( x − 5) = x + 5
=∫ dx
( x − 5)( x + 2)

26. x+5 x+5
A B
( iii ) ∫ 2 dx + =
( x − 5) ( x + 2) ( x − 5)( x + 2)
x − 3 x − 10
x+5
A( x + 2 ) + B( x − 5) = x + 5
=∫ dx
( x − 5)( x + 2) x = −2
− 7B = 3
−3
B=
7

27. x+5 x+5
A B
( iii ) ∫ 2 dx + =
( x − 5) ( x + 2) ( x − 5)( x + 2)
x − 3 x − 10
x+5
A( x + 2 ) + B( x − 5) = x + 5
=∫ dx
( x − 5)( x + 2) x = −2 x=5
− 7B = 3 7 A = 10
−3 10
B= A=
7 7

28. x+5 x+5
A B
( iii ) ∫ 2 dx + =
( x − 5) ( x + 2) ( x − 5)( x + 2)
x − 3 x − 10
x+5
A( x + 2 ) + B( x − 5) = x + 5
=∫ dx
( x − 5)( x + 2) x = −2 x=5
 10 3 − 7B = 3 7 A = 10
= ∫ −  dx
 7 ( x − 5 ) 7( x + 2 )  −3 10
B= A=
7 7

29. x+5 x+5
A B
( iii ) ∫ 2 dx + =
( x − 5) ( x + 2) ( x − 5)( x + 2)
x − 3 x − 10
x+5
A( x + 2 ) + B( x − 5) = x + 5
=∫ dx
( x − 5)( x + 2) x = −2 x=5
 10 3 − 7B = 3 7 A = 10
= ∫ − dx

 7 ( x − 5 ) 7( x + 2 )  −3 10
B= A=
10 3
= log( x − 5) − log( x + 2 ) + c 7 7
7 7

30. x+5 x+5
A B
( iii ) ∫ 2 dx + =
( x − 5) ( x + 2) ( x − 5)( x + 2)
x − 3 x − 10
x+5
A( x + 2 ) + B( x − 5) = x + 5
=∫ dx
( x − 5)( x + 2) x = −2 x=5
 10 3 − 7B = 3 7 A = 10
= ∫ − dx

 7 ( x − 5 ) 7( x + 2 )  −3 10
B= A=
10 3
= log( x − 5) − log( x + 2 ) + c 7 7
7 7
dx
( iv ) ∫ 3
x +x

31. x+5 x+5
A B
( iii ) ∫ 2 dx + =
( x − 5) ( x + 2) ( x − 5)( x + 2)
x − 3 x − 10
x+5
A( x + 2 ) + B( x − 5) = x + 5
=∫ dx
( x − 5)( x + 2) x = −2 x=5
 10 3 − 7B = 3 7 A = 10
= ∫ − dx

 7 ( x − 5 ) 7( x + 2 )  −3 10
B= A=
10 3
= log( x − 5) − log( x + 2 ) + c 7 7
7 7
dx
( iv ) ∫ 3
x +x
dx
=∫
x( x 2 + 1)

32. x+5 x+5
A B
( iii ) ∫ 2 dx + =
( x − 5) ( x + 2) ( x − 5)( x + 2)
x − 3 x − 10
x+5
A( x + 2 ) + B( x − 5) = x + 5
=∫ dx
( x − 5)( x + 2) x = −2 x=5
 10 3 − 7B = 3 7 A = 10
= ∫ − dx

 7 ( x − 5 ) 7( x + 2 )  −3 10
B= A=
10 3
= log( x − 5) − log( x + 2 ) + c 7 7
7 7
A Bx + C 1
dx +2 =
( iv ) ∫ 3 x x + 1 x( x 2 + 1)
x +x
A( x 2 + 1) + ( Bx + C ) x = 1
dx
=∫
x( x 2 + 1)

33. x+5 x+5
A B
( iii ) ∫ 2 dx + =
( x − 5) ( x + 2) ( x − 5)( x + 2)
x − 3 x − 10
x+5
A( x + 2 ) + B( x − 5) = x + 5
=∫ dx
( x − 5)( x + 2) x = −2 x=5
 10 3 − 7B = 3 7 A = 10
= ∫ − dx

 7 ( x − 5 ) 7( x + 2 )  −3 10
B= A=
10 3
= log( x − 5) − log( x + 2 ) + c 7 7
7 7
A Bx + C 1
dx +2 =
( iv ) ∫ 3 x x + 1 x( x 2 + 1)
x +x
A( x 2 + 1) + ( Bx + C ) x = 1
dx
=∫
x( x 2 + 1)
x=0
A =1

34. x+5 x+5
A B
( iii ) ∫ 2 dx + =
( x − 5) ( x + 2) ( x − 5)( x + 2)
x − 3 x − 10
x+5
A( x + 2 ) + B( x − 5) = x + 5
=∫ dx
( x − 5)( x + 2) x = −2 x=5
 10 3 − 7B = 3 7 A = 10
= ∫ − dx

 7 ( x − 5 ) 7( x + 2 )  −3 10
B= A=
10 3
= log( x − 5) − log( x + 2 ) + c 7 7
7 7
A Bx + C 1
dx +2 =
( iv ) ∫ 3 x x + 1 x( x 2 + 1)
x +x
A( x 2 + 1) + ( Bx + C ) x = 1
dx
=∫
x( x 2 + 1) x=i
x=0
− B + Ci = 1
A =1

35. x+5 x+5
A B
( iii ) ∫ 2 dx + =
( x − 5) ( x + 2) ( x − 5)( x + 2)
x − 3 x − 10
x+5
A( x + 2 ) + B( x − 5) = x + 5
=∫ dx
( x − 5)( x + 2) x = −2 x=5
 10 3 − 7B = 3 7 A = 10
= ∫ − dx

 7 ( x − 5 ) 7( x + 2 )  −3 10
B= A=
10 3
= log( x − 5) − log( x + 2 ) + c 7 7
7 7
A Bx + C 1
dx +2 =
( iv ) ∫ 3 x x + 1 x( x 2 + 1)
x +x
A( x 2 + 1) + ( Bx + C ) x = 1
dx
=∫
x( x 2 + 1) x=i
x=0
− B + Ci = 1
A =1
B = −1 C = 0

36. x+5 x+5
A B
( iii ) ∫ 2 dx + =
( x − 5) ( x + 2) ( x − 5)( x + 2)
x − 3 x − 10
x+5
A( x + 2 ) + B( x − 5) = x + 5
=∫ dx
( x − 5)( x + 2) x = −2 x=5
 10 3 − 7B = 3 7 A = 10
= ∫ − dx

 7 ( x − 5 ) 7( x + 2 )  −3 10
B= A=
10 3
= log( x − 5) − log( x + 2 ) + c 7 7
7 7
A Bx + C 1
dx +2 =
( iv ) ∫ 3 x x + 1 x( x 2 + 1)
x +x
A( x 2 + 1) + ( Bx + C ) x = 1
dx
=∫
x( x 2 + 1) x=i
x=0
 1 − x  dx
= ∫ − B + Ci = 1
A =1

 x x + 1
2
B = −1 C = 0

37. x+5 x+5
A B
( iii ) ∫ 2 dx + =
( x − 5) ( x + 2) ( x − 5)( x + 2)
x − 3 x − 10
x+5
A( x + 2 ) + B( x − 5) = x + 5
=∫ dx
( x − 5)( x + 2) x = −2 x=5
 10 3 − 7B = 3 7 A = 10
= ∫ − dx

 7 ( x − 5 ) 7( x + 2 )  −3 10
B= A=
10 3
= log( x − 5) − log( x + 2 ) + c 7 7
7 7
A Bx + C 1
dx +2 =
( iv ) ∫ 3 x x + 1 x( x 2 + 1)
x +x
A( x 2 + 1) + ( Bx + C ) x = 1
dx
=∫
x( x 2 + 1) x=i
x=0
 1 − x  dx
= ∫ − B + Ci = 1
A =1

 x x + 1
2
B = −1 C = 0
= log x − log( x 2 + 1) + c
1
2

38. xdx
( v) ∫
( x + 1) 2 ( x 2 + 1)

39. xdx Cx + D
A B x
( v) ∫ + +2 =
( x + 1) 2 ( x 2 + 1) ( x + 1) ( x + 1) ( x + 1) ( x + 1) 2 ( x 2 + 1)
2
A( x + 1)( x + 1) + B ( x + 1) + ( Cx + D )( x + 1)
2
=x
2 2

40. xdx Cx + D
A B x
( v) ∫ + +2 =
( x + 1) 2 ( x 2 + 1) ( x + 1) ( x + 1) ( x + 1) ( x + 1) 2 ( x 2 + 1)
2
A( x + 1)( x + 1) + B ( x + 1) + ( Cx + D )( x + 1)
2
=x
2 2
x = −1
2 B = −1
−1
B=
2

41. xdx Cx + D
A B x
( v) ∫ + +2 =
( x + 1) 2 ( x 2 + 1) ( x + 1) ( x + 1) ( x + 1) ( x + 1) 2 ( x 2 + 1)
2
A( x + 1)( x + 1) + B ( x + 1) + ( Cx + D )( x + 1)
2
=x
2 2
x=i
x = −1
− 2C + 2 Di = i
2 B = −1
−1
B=
2

42. xdx Cx + D
A B x
( v) ∫ + +2 =
( x + 1) 2 ( x 2 + 1) ( x + 1) ( x + 1) ( x + 1) ( x + 1) 2 ( x 2 + 1)
2
A( x + 1)( x + 1) + B ( x + 1) + ( Cx + D )( x + 1)
2
=x
2 2
x=i
x = −1
− 2C + 2 Di = i
2 B = −1
1
C =0 D=
−1
B= 2
2

43. xdx Cx + D
A B x
( v) ∫ + +2 =
( x + 1) 2 ( x 2 + 1) ( x + 1) ( x + 1) ( x + 1) ( x + 1) 2 ( x 2 + 1)
2
A( x + 1)( x + 1) + B ( x + 1) + ( Cx + D )( x + 1)
2
=x
2 2
x=i
x = −1
− 2C + 2 Di = i
2 B = −1
1
−1 C =0 D=
B= 2
2
x=0
2A + B + D = 0
11
2A − + = 0
22
A=0

44. xdx Cx + D
A B x
( v) ∫ + +2 =
( x + 1) 2 ( x 2 + 1) ( x + 1) ( x + 1) ( x + 1) ( x + 1) 2 ( x 2 + 1)
2
A( x + 1)( x + 1) + B ( x + 1) + ( Cx + D )( x + 1)
2
=x
2 2
x=i
x = −1
 −1 1
= ∫ +  dx
 2( x + 1) 2( x + 1) 
− 2C + 2 Di = i
2 B = −1
2 2
1
−1 C =0 D=
B= 2
2
x=0
2A + B + D = 0
11
2A − + = 0
22
A=0

45. xdx Cx + D
A B x
( v) ∫ + +2 =
( x + 1) 2 ( x 2 + 1) ( x + 1) ( x + 1) ( x + 1) ( x + 1) 2 ( x 2 + 1)
2
A( x + 1)( x + 1) + B ( x + 1) + ( Cx + D )( x + 1)
2
=x
2 2
x=i
x = −1
 −1 1
= ∫ +  dx
 2( x + 1) 2( x + 1) 
− 2C + 2 Di = i
2 B = −1
2 2
1
−1 C =0 D=
B=
1 1 2
− ( x + 1) + 2
∫
−2
=  dx 2
( x + 1) 
2 x=0
2A + B + D = 0
11
2A − + = 0
22
A=0

46. xdx Cx + D
A B x
( v) ∫ + +2 =
( x + 1) 2 ( x 2 + 1) ( x + 1) ( x + 1) ( x + 1) ( x + 1) 2 ( x 2 + 1)
2
A( x + 1)( x + 1) + B ( x + 1) + ( Cx + D )( x + 1)
2
=x
2 2
x=i
x = −1
 −1 1
= ∫ +  dx
 2( x + 1) 2( x + 1) 
− 2C + 2 Di = i
2 B = −1
2 2
1
−1 C =0 D=
B=
1 1 2
− ( x + 1) + 2
∫
−2
=  dx 2
( x + 1) 
2 x=0
 − ( x + 1) −1 −1  2A + B + D = 0
1
= + tan x  + c
2  −1  11
2A − + = 0
22
A=0

47. xdx Cx + D
A B x
( v) ∫ + +2 =
( x + 1) 2 ( x 2 + 1) ( x + 1) ( x + 1) ( x + 1) ( x + 1) 2 ( x 2 + 1)
2
A( x + 1)( x + 1) + B ( x + 1) + ( Cx + D )( x + 1)
2
=x
2 2
x=i
x = −1
 −1 1
= ∫ +  dx
 2( x + 1) 2( x + 1) 
− 2C + 2 Di = i
2 B = −1
2 2
1
−1 C =0 D=
B=
1 1 2
− ( x + 1) + 2
∫
−2
=  dx 2
( x + 1) 
2 x=0
 − ( x + 1) −1 −1  2A + B + D = 0
1
= + tan x  + c
2  −1  11
2A − + = 0
22
1 1 
+ tan −1 x  + c
= A=0
2  x +1 

48. xdx Cx + D
A B x
( v) ∫ + +2 =
( x + 1) 2 ( x 2 + 1) ( x + 1) ( x + 1) ( x + 1) ( x + 1) 2 ( x 2 + 1)
2
A( x + 1)( x + 1) + B ( x + 1) + ( Cx + D )( x + 1)
2
=x
2 2
x=i
x = −1
 −1 1
= ∫ +  dx
 2( x + 1) 2( x + 1) 
− 2C + 2 Di = i
2 B = −1
2 2
1
−1 C =0 D=
B=
1 1 2
− ( x + 1) + 2
∫
−2
=  dx 2
( x + 1) 
2 x=0
 − ( x + 1) −1 −1  2A + B + D = 0
1
= + tan x  + c
2  −1  11
2A − + = 0
22
Exercise 2G;
1 1 
+ tan −1 x  + c
= A=0
1, 3, 5, 7 to 21
2  x +1 