Partial Fraction Decomposition
SUGGESTED REFERENCE MATERIAL:
As you work through the problems listed below, you should reference Chapter 7.5 of the recommended textbook (or the equivalent chapter in your alternative textbook/online resource)
and your lecture notes.
EXPECTED SKILLS:
• Be able to recognize an improper rational function, and perform the necessary long
division to turn it into a proper rational function.
• Know how to write down the partial fraction decomposition for a proper rational
function, compute the unknown coefficients in the partial fractions, and integrate each
partial fraction.
PRACTICE PROBLEMS:
For problems 1-3, write out the partial fraction decomposition. (Do not solve
for the numerical values of the coefficients.)
1.
2x + 3
(x − 2)(x − 5)
B
A
+
x−2 x−5
2.
6
x2 (x2
− 9)
C
D
A B
+ 2+
+
x x
x−3 x+3
3.
5x4 − 1
x(x − 2)(x2 + x + 1)2
A
B
Cx + D
Ex + F
+
+ 2
+ 2
x x − 2 x + x + 1 (x + x + 1)2
For problems 4 & 5, use the given partial fraction decomposition to evaluate the
integral.
Z
11x2 − 28x + 20
4.
dx
(2x + 1)(x − 3)2
1
Hint:
4
5
11x2 − 28x + 20
3
+
+
=
2
(2x + 1)(x − 3)
2x + 1 x − 3 (x − 3)2
5
3
ln |2x + 1| + 4 ln |x − 3| −
+C
2
(x − 3)
Z
5.
11x2 + 7x − 15
dx
(3x − 4)(x2 + 1)
Hint:
11x2 + 7x − 15
5
2x + 5
=
+
(3x − 4)(x2 + 1)
3x − 4 x2 + 1
5
ln |3x − 4| + ln(x2 + 1) + 5 tan−1 x + C
3
For problems 6-17, evaluate the given integral.
Z
4
6.
dx
2
x −1
2 ln |x − 1| − 2 ln |x + 1| + C
Z
4x − 1
dx
7.
2
x − 5x + 6
11 ln |x − 3| − 7 ln |x − 2| + C
Z
8.
x2
dx
x2 + 1
x − tan−1 x + C
Z
3x2 − 4
9.
dx
x+1
3 2
x − 3x − ln |x + 1| + C
2
Z
4x − 1
10.
dx
2
2x − 18x + 36
23
11
ln |x − 6| −
ln |x − 3| + C
6
6
Z
1
11.
dx
x2 (x − 1)2
2 ln |x| −
1
1
− 2 ln |x − 1| −
+ C; Detailed Solution: Here
x
(x − 1)
2
Z
12.
2x + 3
dx
(x − 3)(x + 1)2
9
1
9
ln |x − 3| −
ln |x + 1| +
+C
16
16
4(x + 1)
Z 4
x − 4x2 + 5
13.
dx
x3 − 4x
x2 5
5
5
− ln |x| + ln |x + 2| + ln |x − 2| + C
2
4
8
8
Z 5
x − 3x3 + 6
14.
dx
x3 + x
1 3
x − 4x + 6 ln |x| − 3 ln |x2 + 1| + 4 arctan (x) + C; Detailed Solution: Here
3
Z 3
x − 6x2 + 3x − 17
15.
dx
x2 + 3
x
1 2
1
−1
√
+C
x − 6x + √ tan
2
3
3
Z
2x2
16.
dx
(x − 1)3
2 ln |x − 1| −
Z
17.
4
1
−
+C
x − 1 (x − 1)2
4x2 − 4x + 2
dx
x2 − x
4x − 2 ln |x| + 2 ln |x − 1| + C
For problems 18-19, evaluate the given integral by making substitution that
transforms the problem into integrating a rational function.
Z
sin x
18.
dx
2
cos x + 6 cos x + 5
1
1
ln | cos x + 5| − ln | cos x + 1| + C
4
4
Z
e5x
19.
dx
e4x − 1
ex +
1
1
1
ln |ex − 1| − ln(ex + 1) − tan−1 (ex ) + C
4
4
2
3
20. By the end of this problem, you will know the antiderivatives of sec x. Observe the
following:
Z
Z
1
dx
cos x
Z
cos x
=
dx
cos2 x
Z
cos x
dx
=
1 − sin2 x
sec x dx =
(a) Use the substitution u = sin x to convert the given integral to an integral of a
rational
function.
Z
1
du
1 − u2
(b) Use partial fractions to evaluate your integral
from part (a). Show that the
1 u + 1 +C
antiderivatives have the form ln 2
u − 1
Z
1
1
−
du
2 u−1
1
1
= ln |u + 1| − ln |u − 1| + C
2 2
1 u + 1 +C
= ln 2
u − 1
1
du =
1 − u2
Z
1
2
1
u+1
Notice that substituting back in for u yields:
Z
1 sin x + 1 sec x dx = ln +C
2
sin x − 1 1 (sin x + 1)(sin x + 1) = ln +C
2
(sin x − 1)(sin x + 1) 1 (sin x + 1)2 = ln +C
2
sin2 x − 1 1 (sin x + 1)2 = ln +C
2
cos2 x sin x + 1 +C
= ln cos x = ln |tan x + sec x| + C
4