Section 5.6:
Integration by Substituion
If the integrand has a quantity inside a set of parentheses or under a radical, you typically
perform the integration using a u-substitution. The usual choice for u will be the quantity
inside the parentheses. Be sure to account for the derivative du.
Z
15x2 5x3 + 6
Example 1. Find
4
dx.
Solution. We use a u-substitution with u = 5x3 + 6:
Z
2
3
Z
4
u4 du
15x (5x + 6) dx =
u = 5x3 + 6
2
du = 15x dx
1
= u5 + C
5
=
Z
Example 2. Find
5
1
5x3 + 6 + C
5
(9x − 5)3 dx.
Solution. We use a u-substitution with u = 9x − 5:
Z
(9x − 5)3 dx =
u = 9x − 5
du = 9 dx
1
9
du = dx
Z
u3 · 19 du
1
=
9
Z
1
=
9
=
=
u3 du
1 4
u
4
+C
1 4
u +C
36
1
(9x − 5)4 + C
36
2
SECTION 5.6: INTEGRATION BY SUBSTITUION
Z
x2
Z
e9x−5 dx.
Example 3. Find
Example 4. Find
Z
Example 5.
Find
p
2x3 + 5 dx.
(3x2 + 2)(x3 + 2x + 1)5 dx
SECTION 5.6: INTEGRATION BY SUBSTITUION
Z
Example 6.
Find
(x + 3)
Z
Example 7.
Find
1
dx
(7x + 4)3
Z
Example 8.
Find
p
x2 + 6x + 1 dx
(x2
x+3
dx
+ 6x − 3)5
3
4
SECTION 5.6: INTEGRATION BY SUBSTITUION
Z
Example 9.
Example 10.
9x(x2 − 6)5 dx
Find
Z
(x + 1)(4x2 + 8x − 9)3 dx
Z
e x
√ dx
x
Find
√
Example 11.
Find
SECTION 5.6: INTEGRATION BY SUBSTITUION
Z
Example 12.
Find
Z
Example 13.
Find
(e−x + 5)3
dx
ex
4x − 3
dx
− 3x + 7
2x2
Z
Example 14.
Find
4x2
x−2
dx
− 16x + 3
5
6
SECTION 5.6: INTEGRATION BY SUBSTITUION
Z
Example 15.
3
Find
(x − 1)3 dx
−2
Z
Example 16.
1
Find
x
p
x2 + 8 dx
−1
Z
Example 17.
Find
3
4
1
dx
2−x
SECTION 5.6: INTEGRATION BY SUBSTITUION
7
EXERCISES
Find the following integrals. Check your answer by differentiating.
Z
1.
Z
2.
Z
2x(x2 + 3)4 dx
2
3
14.
5
(3x + 2)(x + 2x + 1) dx
Z
8
(4x − 5) dx
3.
Z
√
Z
15.
Z
16.
5x + 3 dx
17.
Z
p
x x2 + 4 dx
18.
Z
p
(x + 2) x2 + 4x + 1 dx
Z
1
dx
(5x − 2)3
4.
5.
1
√
dx
3
4x + 1
(x + 1)(3x2 + 6x − 1)3 dx
√
1
dx
3x + 5
Z
e4x−3 dx
Z
xex dx
Z
e x
√ dx
x
Z
(e−x + 2)6
dx
ex
2
√
6.
7.
Z
8.
Z
9.
Z
10.
x
dx
2
(x + 6)3
x+3
dx
2
(x + 6x − 3)5
2x(3x2 − 1)4 dx
19.
20.
Z
0
(x + 1)3 dx
21.
−1
Z
2
(x − 2)2 dx
22.
0
Z
23.
4√
2x + 1 dx
0
Z
11.
5x(x2 + 3)5 dx
Z
1
24.
x
p
x2 + 3 dx
−1
Z
12.
√
3
8x − 1 dx
Z
1
x2 (x3 + 2)2 dx
25.
0
Z
13.
x2 (x3 − 6)4 dx
Z
26.
0
2
x2
p
x3 + 1 dx
8
SECTION 5.6: INTEGRATION BY SUBSTITUION
ANSWERS
1.
1 2
(x + 3)5 + C
5
14.
3
(4x + 1)2/3 + C
8
2.
1 3
(x + 2x + 1)6 + C
6
15.
1
(3x2 + 6x − 1)4 + C
24
3.
1
(4x − 5)9 + C
36
16.
2
(3x + 5)1/2 + C
3
4.
2
(5x + 3)3/2 + C
15
17.
1 4x−3
e
+C
4
5.
1 2
(x + 4)3/2 + C
3
18.
1 x2
e +C
2
6.
1 2
(x + 4x + 1)3/2 + C
3
19.
2e
7.
−
1
(5x − 2)−2 + C
10
20.
1
− (e−x + 2)7 + C
7
8.
1
− (x2 + 6)−2 + C
4
21.
1/4
9.
1
− (x2 + 6x − 3)−4 + C
8
22.
8/3
10.
1
(3x2 − 1)5 + C
15
23.
26/3
11.
5 2
(x + 3)6 + C
12
24.
0
12.
3
(8x − 1)4/3 + C
32
25.
19/9
13.
1 3
(x − 6)5 + C
15
26.
52/9
√
x
+C