Integration:
Integration by Partial Fractions
Step 1 If you are integrating a rational function p(x)
where degree of p(x) is greater than degree of
q(x)
q(x), divide the denominator into the numerator, then proceed to the step 2 and then 3a or 3b
or 3c or 3d followed by Step 4 and Step 5.
Z
x2 − 5x + 7
dx =
x2 − 5x + 6
Z 1
1+ 2
x − 5x + 6
Z
dx =
Z
dx +
x2
dx
− 5x + 6
Step 2 Factor denominator into irreducible factors
Z
Z
dx
dx
=
2
x − 5x + 6
(x − 3)(x − 2)
Step 3a If factors are linear put in form
Z
A
(x+c)
+
B
(x+d)
dx
=
(x − 3)(x − 2)
Z
and find A and B .
A
dx +
x−3
Z
B
dx
x−2
Step 3b If factors are linear but squared put in form:
Z
Z
Z
Z
A
B
C
dx
=
dx +
dx +
dx
2
(x − 2)(x + 1)
x−2
x+1
(x + 1)2
Step 3c If factors are quadratic put in form:
Z
Z
Z
dx
A
Bx + C
=
dx +
dx
2
2
(x − 1)(x − 3x − 2)
x−1
x − 3x − 2
Step 3d If factors are quadratics but squared put in form:
Z
R A
R Bx+C
dx
=
dx
+
dx
x−1
x2 −3x−2
(x − 1)(x2 − 3x − 2)2
R
dx
+ (x2Dx+E
−3x−2)2
Step 4 Get common denominator on right hand side and equate coefficients:
Z
R A
R B
R C
dx
=
dx
+
dx
+
dx
x−2
x+1
(x+1)2
(x − 2)(x + 1)2
R A(x+1)2 +B(x−2)(x+1)+C(x−2)
dx
=
(x−2)(x+1)2
R (A+B)x2 +(2A−B+C)x+(A−2B−2C)
=
dx
(x−2)(x+1)2

A+B =0



2A − B + C = 0
⇒
A − 2B − 2C = 1



Z
⇒
⇒ A = −B
(equating coefficients of x2 )
⇒ A = −C/3 (equating coefficients of x)
⇒ A = 1/9, B = −1/9, C = −1/3
(equating numbers)
dx
=
(x − 2)(x + 1)2
Z
1
dx −
9(x − 2)
Z
1
dx −
9(x + 1)
Z
1
dx
3(x + 1)2
Step 5 Integrate by substitution.
Let
u = x − 2 ⇒ du = dx,
v = x + 1 ⇒ dv = dx
Z
⇒
Z
Z
Z
1
du 1
dv 1
dx
=
−
−
v −2 dv
(x − 2)(x + 1)2
9
u
9
v
3
1
1
1
=
ln |u| − ln |v| + v + c
9
9
3
1
1
1
=
ln |x − 2| − ln |x + 1| + (x + 1) + c,
9
9
3
where c = constant.
Examples to try
x − 1
9
+ 3 +c
dx
=
ln
x + 2 x + 2
(x − 1)(x + 2)2
Z
2 + 3x + x2
1
2
−1
dx
=
2
ln
|x|
−
ln
|x
+
1|
+
3
tan
x+c
x(x2 + 1)
2
Z
x2 x2 + 2
3
+
dx
=
ln
2x2 + 1 4(2x2 + 1) + c
4x5 + 4x3 + x
Z