1
Worksheet 12, Math 10560
The partial fraction decomposition of the rational function
2x + 1
(x + 1)2 (x2 + 4)3
is of the form:
A
Bx + C
Cx + D Ex3 + F x2 + Gx + H Ix5 + Jx4 + Kx3 + Lx2 + M x + N
+
+
+ 2
+
x + 1 (x + 1)2
x +4
(x2 + 4)2
(x2 + 4)3
Ax + B
Cx + D
+
(x + 1)2 (x2 + 4)3
A
B
Ex + F
Cx + D
Gx + H
+
+ 2
+ 2
+ 2
2
2
x + 1 (x + 1)
x +4
(x + 4)
(x + 4)3
Cx5 + Dx4 + Ex3 + F x2 + Gx + H
Ax + B
+
(x + 1)2
(x2 + 4)3
A
B
Cx + D
Gx + H
+
+
+
x + 1 (x + 1)2
x2 + 4
(x2 + 4)3
Solution: The denominator is the product of a repeated linear factor (x + 1)2 and a repeated
irreducible quadratic factor (x2 + 4)2 . Our decomposition must be of the form
A
B
Cx + D
Ex + F
Gx + H
+
+
+
+
.
x + 1 (x + 1)2
x2 + 4
(x2 + 4)2 (x2 + 4)3
2
Which of the following expressions give the partial fraction decomposition of the function
f (x) =
4x2 + 3x + 8
.
x(x − 2)3 (x2 + 8)
A
B
C
D
Ex + F
+
+
+
+ 2
2
3
x x − 2 (x − 2)
(x − 2)
x +8
A Bx + C Dx + E
F x + G Hx + I
+
+
+
+ 2
2
x
x−2
(x − 2)
(x − 2)3
x +8
A Bx2 + Cx + D Ex + F
+
+ 2
x
(x − 2)3
x +8
A
B
Cx + D
+
+ 2
x x−2
x +8
A
B
C
D
E
+
+
+
+ 2
2
3
x x − 2 (x − 2)
(x − 2)
x +8
Solution:
Page 2
3
Find
Z
2x3 − 2x2 + 1
dx
x2 − x
Solution:
4
Give the FORM of the partial fraction. Do NOT solve for the coefficients.
3x2 − 9x
=
x(x − 1)2 (x2 + 1)2
Solution:
Page 3
5
Calculate the integral
Z
dx
√ .
x+ 3x
Solution: Make substitution u = x1/3 . Then u3 = x and with dx = 3u2 du
Z
Z
Z
3udu
dx
3
3
3u2 du
√
=
= ln(u2 + 1) = ln(x2/3 + 1) + C.
=
3
2
2
u(u + 1)
u +1
2
2
x+ x
6
Evaluate the integral
Z
1
(1 −
√ 8
x) dx.
0
√
√
Solution: We do this with u-substitution. Let u = 1− x so that x = 1−u and hence x = (1−u)2 .
Using this, we see that dx = −2(1 − u)du. Also, the bounds of integration go from x = 0 to u = 1
and from x = 1 to u = 0. Making the substitution gives:
Z 0
Z 1
Z 1
√ 8
8
u8 − u9 du
−2(1 − u)u du = 2
(1 − x) dx =
1
0
0
1
9
u
u10 1
1
1
1
=2
−
−
−0 =2
=
=2
9
10 9 10
90
45
0
Page 4
7
Find
Solution: Let u =
Z
x2 + 3x
√
dx.
x+4
√
x + 4. then du = 21 (x + 4)−1/2 dx and x = u2 − 4. Hence dx = 2udu and
Z
x2 + 3x
√
dx = 2((u2 − 4)2 + 3(u2 − 4)) du
x+4
Z
5u3 u5
= 2 4 − 5u2 + u4 du = 2[4u −
+ ]+C
3
5
h √
(x + 4)5/2 i
5
3/2
= 2 4 x + 4 − (x + 4) +
+ C.
3
5
2√
=
x + 4 (8 + x + 3x2 ) + C.
15
Z
8
Find
Solution: Let u =
Z
√
x
dx.
x+1
√
x. then du = 21 (x)−1/2 dx and x = u2 . Hence
Z
Z
=
√
x
dx. =
x+1
Z
2u2
du
u2 + 1
2
du = 2u − 2 tan−1 (u) + C
+1
√
√
= 2 x − 2 tan−1 ( x + C.
2−
u2
Page 5
Z
9
Find
√
e
x
dx
Solution:
10
Compute the integral
Z
2
dx .
+ 1)
x(x2
Solution: We write
A Bx + C
2
= + 2
+ 1)
x
x +1
x(x2
Then grouping by powers of x, we determine
x2 :
x1 :
x0 :
A+B =0
C=0
A=2
Hence, A = 2, B = −2, and C = 0. Therefore,
R
R 2
R
2
dx =
dx −
x(x2 +1)
x
2x
dx
x2 +1
= ln x2 − ln(x2 + 1) + C
Page 6
11
Evaluate the integral
Z
Solution:
4x2 + 3
dx.
x3 + x
4x2 + 3
A Bx + C
4x2 + 3
=
= + 2
3
2
x +x
x(x + 1)
x
x +1
Multiplying both sides by x(x2 + 1), we get
4x2 + 3 = A(x2 + 1) + (Bx + C)x = Ax2 + Bx2 + Cx + A
Comparing coefficients, we get
→
A = 3, C = 0, A + B = 4,
Hence we have
and
A = 3, C = 0, B = 1.
4x2 + 3
3
x
= + 2
3
x +x
x x +1
Z
4x2 + 3
dx =
x3 + x
Z
Z
3
x
dx +
dx
2
x
x +1
Z
1
1
= 3 ln |x| +
du
2
u
(where u = x2 + 1)
== 3 ln |x| +
== 3 ln |x| +
1
ln |u| + C
2
1
ln |x2 + 1| + C
2
Page 7
12
Evaluate the integral
Z
x3 + 4x + 6
dx.
x4 + 4x2
Solution: Use a partial fraction decomposition. Factoring the denominator
x4 + 4x2 = x2 (x2 + 4)
where x2 is a repeated linear factor and x2 + 4 is an irreducible quadratic factor. Therefore the
decomposition is of the form
x3 + 4x + 6
A B
Cx + D
= + 2+ 2
2
2
x (x + 4)
x x
x +4
This leads to the equation:
x3 + 4x + 6 = Ax(x2 + 4) + B(x2 + 4) + (Cx + D)x2 = Ax3 + 4Ax + Bx2 + 4B + Cx3 + Dx2
which breaks down into system of equations
A+C =1
B+D =0
4A = 4
4B = 6.
We solve this to find A = 1, B = 23 , C = 0, and D = − 32 . Therefore
Z
x3 + 4x + 6
dx =
x4 + 4x2
Z
Z
Z
1
3
1
3
1
dx +
dx −
dx
2
2
x
2
x
2
x +4
3
3
x
= ln |x| −
− arctan + C.
2x 4
2
Page 8
Z
13
Evaluate
0
1
2x2
dx
(x + 1)(x2 + 1)
Solution:
14
Find the integral
Z
3x + 1
dx.
x3 + x2
Solution: Use partial fraction decomposition
3x + 1
A B
C
Ax(x + 1) + B(x + 1) + Cx2
3x + 1
=
=
+
+
=
.
x 3 + x2
x2 (x + 1)
x x2 x + 1
x2 (x + 1)
Therefore
3x + 1 = (A + C)x2 + (A + B)x + B.
It follows that
A + C = 0,
A = 2,
and
Z
3x + 1
dx =
x3 + x2
Z
A + B = 3,
B = 1,
B = 1,
C = −2,
2
1
2
1
( + 2−
)dx = 2 ln |x| − − 2 ln |x + 1| + C.
x x
x+1
x
Page 9