Raney
Calculus I
Summer 2019
Solutions to Worksheet 22
1. Use a substitution to evaluate the indefinite integral.
Z
(a) (2x + 5)7 dx
Let u = 2x + 5. Then du = 2 dx, and so dx =
Z
Z
(b)
x
p
1
2
du. After substituting we get
Z
1
=
u7
du
2
1 1 8
=
· u +C
2 8
1
(2x + 5)8 + C
=
16
(2x + 5)7 dx
x2 + 2 dx
Let u = x2 + 2. Then du = 2x dx, and so x dx =
Z
x
p
x2 + 2 dx
=
=
=
=
=
Z
(c)
1
2
du. After substituting we get
Z
√
1
u
du
2
Z
1
u1/2 du
2
1 1 3/2
·
u +C
2 3/2
1 2 2
· (x + 2)3/2 + C
2 3
1 2
(x + 2)3/2 + C
3
x2 (x3 − 1)4 dx
Let u = x3 − 1. Then du = 3x2 dx, and so x2 dx =
Z
x2 (x3 − 1)4 dx
=
=
(d)
du. After substituting we get
1
4
u
du
3
1 1 5
· u +C
3 5
1 3
(x − 1)5 + C
15
Z
=
Z
1
3
x
dx
1 − x2
Let u = 1 − x2 . Then du = −2x dx, and so x dx = − 21 du. After substituting we get
√
Z
√
x
dx
1 − x2
− 21 du
√
u
Z
1
−
u−1/2 du
2
1 1 1/2
− ·
u +C
2 1/2
1
− · 2(1 − x2 )1/2 + C
2
p
− 1 − x2 + C
Z
=
=
=
=
=
4x3 + 3
dx
x4 + 3x
Let u = x4 + 3x. Then du = (4x3 + 3) dx. After substituting we get
Z
Z
4x3 + 3
1
dx
=
du
x4 + 3x
u
= ln |u| + C
Z
(e)
ln |4x3 + 3| + C
=
(ln x)2
dx
x
Let u = ln x. Then du =
Z
(f)
1
x
dx. After substituting we get
Z
(ln x)2
dx
x
Z
1 3
u +C
3
1
(ln x)3 + C
3
=
=
Z
(g)
4
x3 ex dx
Let u = x4 . Then du = 4x3 dx, and so x3 dx =
Z
Z
(h)
u2 dx
=
1
4
du. After substituting we get
Z
1
u
du
=
e
4
1 u
=
e +C
4
1 x4
e +C
=
4
4
x3 ex dx
x
dx
1 − x4
We note that
√
Z
x
√
dx
1 − x4
Z
=
Let u = x2 . Then du = 2x dx, and so x dx =
Z
x
p
dx
1 − (x2 )2
x
p
1
2
1 − (x2 )2
dx
du. After substituting we get
Z
=
=
=
1
du
√2
1 − u2
1
sin−1 u + C
2
1
sin−1 (x2 ) + C
2
2. Use a substitution to evaluate the definite integral.
Z 1
(a)
x(x2 + 1)3 dx
−1
Let u = x2 + 1. Then du = 2x dx, and so x dx = 12 du.
When x = −1, u = (−1)2 + 1 = 1 + 1 = 2, and when x = 1, u = 12 + 1 = 1 + 1 = 2.
So after the substitution we get
Z 2 Z 1
1
u3
x(x2 + 1)3 dx =
du
2
2
−1
This equals 0 since the new limits of integration are the same.
Z 4
1
√
(b)
dx
2x + 1
0
Let u = 2x + 1. Then du = 2 dx, and so dx = 12 du.
When x = 0, u = 2(0) + 1 = 1, and when x = 4, u = 2(4) + 1 = 8 + 1 = 9.
So after the substitution we get
Z 4
Z 9
1
1
1
√
√
du
dx =
u 2
2x + 1
0
1
Z 9
1
=
u−1/2 du
2 1
9
=
=
Z
(c)
1
1 1 1/2
·
u
2 1/2
1
√
√
√ 9
u1 =
9− 1 = 3−1 = 2
2
xe−x dx
0
Let u = −x2 . Then du = −2x dx, and so x dx = − 12 du.
When x = 0, u = −02 = −0 = 0, and when x = 1, u = −12 = −1.
So after the substitution we get
Z 1
Z −1 2
1
xe−x dx =
eu − du
2
0
0
Z −1
1
= −
eu du
2 0
−1
1
− eu
2 0
1 −1
1 0
= − e − − e
2
2
1 1 1
1
1
= − · + ·1 =
1−
2 e 2
2
e
=
Z
(d)
e
e2
1
dx
x ln x
1
dx.
x
When x = e, u = ln e = 1, and when x = e2 , u = ln e2 = 2 ln e = 2.
So after the substitution we get
Let u = ln x. Then du =
Z
e
Z
(e)
e2
1
dx
x ln x
Z
2
=
1
du
1 u
u
ln |u| |1
=
ln 2 − ln 1 = ln 2 − 0 = ln 2
=
π/2
sin x
dx
1
+
cos2 x
0
Let u = cos x. Then du = − sin x dx, and so sin x dx = −du.
When x = 0, u = cos 0 = 1, and when x = π/2, u = cos(π/2) = 0.
So after the substitution we get
Z
0
π/2
sin x
dx
1 + (cos x)2
0
−du
1 + u2
Z
0
Z
=
1
= −
1
=
1
du
1 + u2
− tan−1 u
0
1
= − tan−1 0 − (− tan−1 1)
π
π
= −0 +
=
4
4