05 - Integration Substitution Power Rule - Kuta Software

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Kuta Software - Infinite Calculus
Name___________________________________
Integration by Substitution
Evaluate each indefinite integral. Use the provided substitution.
Date________________ Period____
1)
∫
−15x 4 (−3x 5 − 1) 5 dx; u = −3x 5 − 12)
∫
−16x 3 (−4x 4 − 1) −5 dx; u = −4x 4 − 1
3)
∫
−
8x 3
(−2x 4 + 5) 5 dx; u = −2x4 + 5 4)
∫
2
(5x 4 + 5)
3 ⋅ 20x 3 dx; u = 5x 4 + 5
5)
∫
(5 + ln x) 5
dx; u = 5 + ln x 6)
x
∫
4sec 4x ⋅ tan 4x ⋅ sec 4 4x dx; u = sec 4x
7)
∫
36x 3 (3x 4 + 3) 5 dx; u = 3x 4 + 3 8)
∫
x(4x − 1) 4 dx; u = 4x − 1

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Evaluate each indefinite integral.
9)
∫
−9x 2 (−3x 3 + 1) 3 dx10)
∫
12x 3 (3x 4 + 4) 4 dx
11)
∫
−12x 2 (−4x 3 + 2) −3 dx12)
∫
3
(3x 5 − 3)
5 ⋅ 15x 4 dx
13)
∫
(−2x 4 − 4) 4 ⋅ −32x 3 dx
14)
∫
1
(e 4x 5
− 4) ⋅ 8e 4x dx
15)
∫
x(4x + 5) 3 dx 16)
∫
5x
2x + 3 dx

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Kuta Software - Infinite Calculus
Name___________________________________
Integration by Substitution
Evaluate each indefinite integral. Use the provided substitution.
Date________________ Period____
1)
∫
−15x 4 (−3x 5 − 1) 5 dx; u = −3x 5 − 1
2)
∫
−16x 3 (−4x 4 − 1) −5 dx; u = −4x 4 − 1
1
6 (−3x5 − 1) 6 + C
−
1
4(−4x 4 − 1) 4
+ C
3)
∫
−
−
8x 3
(−2x 4 + 5) 5 dx; u = −2x4 + 5
1
4(−2x 4 + 5) 4 + C 4)
∫
2
(5x 4 + 5)
3 ⋅ 20x 3 dx; u = 5x 4 + 5
3
5 (5x4 + 5)
3 + C
5
5)
∫
(5 + ln x) 5
dx; u = 5 + ln x
x
1
6 (5 + ln x)6 + C
6)
∫
4sec 4x ⋅ tan 4x ⋅ sec 4 4x dx; u = sec 4x
1
5 ⋅ sec5 4x + C
7)
∫
36x 3 (3x 4 + 3) 5 dx; u = 3x 4 + 3
1
2 (3x4 + 3) 6 + C
8)
∫
x(4x − 1) 4 dx; u = 4x − 1
1
96 (4x − 1)6 + 1
80 (4x − 1)5 + C

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Evaluate each indefinite integral.
9)
∫
−9x 2 (−3x 3 + 1) 3 dx
1
4 (−3x3 + 1) 4 + C
10)
∫
12x 3 (3x 4 + 4) 4 dx
1
5 (3x4 + 4) 5 + C
11)
∫
−12x 2 (−4x 3 + 2) −3 dx
−
1
2(−4x 3 + 2) 2 + C 12)
∫
3
(3x 5 − 3)
5 ⋅ 15x 4 dx
5
8 (3x5 − 3)
5 + C
8
13)
∫
(−2x 4 − 4) 4 ⋅ −32x 3 dx
4
5 (−2x4 − 4) 5 + C
14)
∫
1
(e 4x 5
− 4) ⋅ 8e 4x dx
5
( e 4x − 4)
3
6
5
+ C
15)
∫
x(4x + 5) 3 dx
1
80 (4x + 5)5 − 5
64 (4x + 5)4 + C
16)
∫
5x
2x + 3 dx
5
1
2 (2x + 3) 2 − 5 3
2 (2x + 3) 2 + C
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