21.
Z
ln(ln x)
dx
x
=
Z
Let u = ln x
dx
du =
x
ln u du
U = ln u
du
dU =
Zu
= u ln u −
d V = du
V =u
du = u ln u − u + C
= (ln x)(ln(ln x)) − ln x + C.
26.
Z
=
(sin−1 x)2 d x
Z
Let x = sin θ
d x = cos θ dθ
θ 2 cos θ dθ
d V = cos θ dθ
U = θ2
dU = 2θ dθ
V = sin θ
Z
= θ 2 sin θ − 2 θ sin θ dθ
U =θ
dU = dθ
d V = sin θ dθ
V = − cos
Zθ
= θ 2 sin θ − 2(−θ cos θ + cos θ dθ )
= θ 2 sin θ + 2θ cos θ − 2 sin θ + C
p
= x(sin−1 x)2 + 2 1 − x 2 (sin−1 x) − 2x + C.
28. By the procedure used in Example 4 of Section 7.1,
Z
Z
e x cos x d x = 12 e x (sin x + cos x) + C;
e x sin x d x = 21 e x (sin x − cos x) + C.
Now
Z
xe x cos x d x
d V = e x cos x d x
V = 21 e x (sin x + cos x)
Z
x
1
1
= 2 xe (sin + cos x) − 2 e x (sin x + cos x) d x
U=x
dU = d x
= 21 xe x (sin + cos x)
− 41 e x (sin x − cos x + sin x + cos x) + C
= 21 xe x (sin x + cos x) − 21 e x sin x + C.
16. First divide to obtain
37x + 85
x3 + 1
= x −7+
2
(x + 4)(x + 3)
x + 7x + 12
A
B
37x + 85
=
+
(x + 4)(x + 3)
x +4 x +3
(A + B)x + 3 A + 4B
=
x 2 + 7x + 12
n
A + B = 37
⇒
⇒ A = 63, B = −26.
3 A + 4B = 85
Now we have
Z x3 + 1
26
63
dx =
−
dx
x −7+
12 + 7x + x 2
x +4 x +3
x2
− 7x + 63 ln |x + 4| − 26 ln |x + 3| + C.
=
2
Z
26. We have
Z
dt
2
2
Z(t − 1)(t − 1)
dt
Let u = t − 1
=
(t − 1)3 (t + 1)2
du = dt
Z
du
=
u 3 (u + 2)2
A
C
D
B
E
1
= + 2+ 3+
+
3
2
u (u + 2)
u
u
u
u + 2 (u + 2)2
A(u 4 + 4u 3 + 4u 2 ) + B(u 3 + 4u 2 + 4u)
=
u 3 (u + 2)2
2
C(u + 4u + 4) + D(u 4 + 2u 3 ) + Eu 3
u 3 (u + 2)2

A+D=0



 4 A + B + 2D + E = 0
⇒ 4 A + 4B + C = 0



 4B + 4C = 0
4C = 1
1
1
3
1
3
, B=− , C = , D=− , E =− .
⇒A=
16
4
4
16
8
Z
du
2
u 3 (u
Z + 2)
Z
Z
1
3
1
du
du
du
+
=
−
16
uZ 4
u2 Z 4
u3
3
1
du
du
−
−
16
u+2 8
(u + 2)2
1
1
3
ln |t − 1| +
−
−
=
16
4(t − 1) 8(t − 1)2
3
1
ln |t + 1| +
+ K.
16
8(t + 1)
27.
Z
e2x
Z
dx
=
− 4e x + 4
Z
(e x
dx
− 2)2
Let u = e x
du = e x d x
du
u(u − 2)2
A
B
C
1
=
+
+
u(u − 2)2
u
u − 2 (u − 2)2
A(u 2 − 4u + 4) + B(u 2 − 2u) + +Cu
=
u(u − 2)2
(
A+B =0
1
1
1
⇒ −4 A − 2B + C = 0 ⇒ A = , B = − , C = .
4
4
2
4A = 1
Z
Z
Z
Z
du
1
1
1
du
du
du
=
−
+
u(u − 2)2
4
u
4
u−2 2
(u − 2)2
1
1
1
1
+K
= ln |u| − ln |u − 2| −
4
4
2 (u − 2)
x
1
1
= − ln |e x − 2| −
+ K.
x
4 4
2(e − 2)
=
40.
Z
dx
− 1)3/2
Let x = sec θ
d x = sec θ tan θ dθ
Z
Z
cos3 θ dθ
sec θ tan θ dθ
=
=
sin2 θ
sec2 θ tan3 θ
Z
2
1 − sin θ
cos θ dθ Let u = sin θ
=
sin2 θ
du = cos θ dθ
Z
1
1 − u2
du = − − u + C
=
2
u
u
1
=−
+ sin θ + C
sin θ
!
√
x
x2 − 1
=− √
+ C.
+
x
x2 − 1
x 2 (x 2
√
x
x 2 −1
θ
1
Fig. 3-40
41.
I =
Z
dx
x(1 + x 2 )3/2
Let x = tan θ
d x = sec2 θ dθ
Z
Z
2
sec θ dθ
cos2 θ dθ
=
=
tan θ sec3 θ
sin θ
Z
2
cos θ sin θ dθ
Let u = cos θ
=
sin2 θ
du = − sin θ dθ
Z 2
Z
u du
du
=−
=u+
.
2
2
1−u
u −1
We have
1
1
1
1
=
−
.
u2 − 1
2 u−1 u+1
Thus
1 u − 1 I = u + ln +C
2
u + 1
1 cos θ − 1 +C
= cos θ + ln 2
cos θ + 1 1
√
− 1
1
1 1 + x2
+C
= √
+ ln 1
2 1 + x2
+ 1 √
1 + x2
!
√
1 + x2 − 1
1
1
+ ln √
+ C.
= √
2
1 + x2
1 + x2 + 1
√
1+x 2
x
θ
1
Fig. 3-41
44.
Z
π/2
dθ
1 + cos θ + sin θ
θ
2
Let x = tan , dθ =
d x,
2
1 + x2
0
1 − x2
2x
cos θ =
, sin θ =
.
2
2
1
+
x
1
+
x
2
dx
Z 1
1 + x2
=
2x
1 − x2
0
+
1+
1 + x2
1 + x2
Z 1
Z 1
dx
dx
=2
=
0 2 + 2x
0 1+x
1
= ln |1 + x| = ln 2.
0
4. As an alternative to the direct method of Example 3, we begin with the change of variable
u = ln x, or, equivalently, x = eu , so that d x = eu du.
I =
Z
2
4
x (ln x) d x =
Z
u 4 e3u du
4
3
2
= a4 u + a3 u + a2 u + a1 u + a0 e3u + C.
We will have
dI
= 4a4 u 3 + 3a3 u 2 + 2a2 u + a1 e3u
du
+ 3 a4 u 4 + a3 u 3 + a2 u 2 + a1 u + a0 e3u
= u 4 e3u
provided 3a4 = 1, 3a3 + 4a4 = 0, 3a2 + 3a3 = 0, 3a1 + 2a2 = 0, and 3a0 + a1 = 0. Thus
a4 = 1/3, a3 = −4/9, a2 = 4/9, a1 = −8/27, and a0 = 8/81. We now have
1 4 4 3 4 2
8
8
u − u + u − u+
e3u + C
3
9
9
27
81
I =
Z
x 2 (ln x)4 d x
4
4(ln x)3
4(ln x)2
8 ln x
8
3 (ln x)
−
+
−
+
+C
=x
3
9
9
27
81
29. Since (x − 1)(x 2 − 1)(x 3 − 1) = (x − 1)3 (x + 1)(x 2 + x + 1), and the numerator has
degree less than the denominator, we have
x5 + x3 + 1
A
B
3
D
Ex + F
=
+
+
+
+
.
(x − 1)(x 2 − 1)(x 3 − 1)
x − 1 (x − 1)2
(x − 1)3
x + 1 x2 + x + 1
19. Suppose that I =
Z
2
2
e−x d x = P(x) e−x + C, where P is a polynomial having, say,
degree m ≥ 0.
(a) Then we must have
dI
2
2
= P ′ (x) − 2x P(x) e−x = e−x .
dx
It follows that P ′ (x) − 2x P(x) = 1. The left side of this equation must be a
polynomial of degree m + 1 ≥ 1 because 2x P(x) has degree m + 1 and P ′ (x)
only has degree m − 1. But the right side of the equation is a polynomial of degree
0 (i.e., a constant). This contradiction shows that no such polynomial P(x) can
exist.
2
d
2
Erf(x) = √ e−x by the Fundamental Theorem of Calculus, we have
(b) Since
dx
π
Z
√
e
−x 2
dx =
π
Erf(x) + C.
2
(c) Let us try the form
J=
Z
2
Erf(x) d x = P(x)Erf(x) + Q(x) e−x + C,
where P and Q are polynomials to be determined. Then
dJ
= P ′ (x) Erf(x)
dx
2
2
′
+ √ P(x) + Q (x) − 2x Q(x) e−x
π
= Erf(x).
2
Hence we must have P ′ (x) = 1 and √ P(x) + Q ′ (x) − 2x Q(x) = 0. The first
π
of these DEs says that P(x) = x + k; without loss of generality we can take the
constant k to be zero. The second DE says that
2x
Q ′ (x) − 2x Q(x) = − √ .
π
The right side has degree 1 and so must√the left side. Thus Q must have degree
zero. Hence Q ′ (x) = 0 and Q(x) = 1/ π . Therefore
J=
Z
1
2
Erf(x) d x = xErf(x) + √ e−x + C.
π
32.
In =
Z
π/2
x n sin x d x
0
d V = sin x d x
U = xn
n−1
dU = nx
dx
V = − cos x
π/2
Z π/2
x n−1 cos x d x
+n
= −x n cos x 0
0
x n−1
d V = cos x d x
U=
n−2
dU = (n − 1)x
dx
V = sin x
π/2
Z π/2
n−2
n−1
x
sin x d x
− (n − 1)
=n x
sin x π n−1
0
0
=n
− n(n − 1)In−2 ,
(n ≥ 2).
2
Z π/2
π/2
I0 =
sin x d x = − cos x = 1.
0
0
h π i
π 5
π 3
− 6(5) 4
− 4(3) 2
I6 = 6
− 2(1)I0
2
2
2
3 5
=
π − 15π 3 + 360π − 720.
16
33.
In =
Z
sinn x d x
(n ≥ 2)
U = sinn−1 x
dU = (n − 1) sinn−2 x cosZ x d x
= − sinn−1 x cos x + (n − 1)
d V = sin x d x
V = − cos x
sinn−2 x cos2 x d x
= − sinn−1 x cos x + (n − 1)(In−2 − In )
n In = − sinn−1 x cos x + (n − 1)In−2
1
n−1
In = − sinn−1 x cos x +
In−2 .
n
n
Note: I0 = x + C, I1 = − cos x + C. Hence
5
1
I6 = − sin5 x cos x + I4
6
6
1 5
5
1 3
3
= − sin x cos x +
− sin x cos x + I2
6
6
4
4
5
1
sin3 x cos x
= − sin5 x cos x −
6
24
5
1
1
+
− sin x cos x + I0
8
2
2
5
5
1
sin3 x cos x −
sin x cos x
= − sin5 x cos x −
6
24
16
5
+ x +C
16
!
5x
sin5 x
5 sin3 x
5 sin x
=
− cos x
+
+
+ C.
16
6
24
16
1
6
I7 = − sin6 x cos x + I5
7
7
6
1 4
4
1 6
− sin x cos x + I3
= − sin x cos x +
7
7
5
5
1 6
6
= − sin x cos x −
sin4 x cos x
7
35
24
2
1 2
+
− sin x cos x + I1
35
3
3
1
6
8
= − sin6 x cos x −
sin4 x cos x −
sin2 x cos x
7
35
35
16
cos x + C
−
35 6
6 sin4 x
8 sin2 x
16
sin x
+
+
+
+ C.
= − cos x
7
35
35
35