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’e’x x 2 ’ 2xe’x ’ 2e’x
N
= lim 2
N ’+∞

’e’N N 2 ’ 2N e’N ’ 2e’N
= lim
N ’+∞

’ ’e’2 22 ’ 2 · 2 · e’2 ’ 2e’2
= e’2 22 + 4e’2 + 2e’2 .
∞ 1 N
x ln x dx = lim x ln x dx + lim
(c) x ln x dx
’0+ N’+∞ 1
0
= lim [x ln x ’ x]1 + lim [x ln x ’ x]N
1
’+ N’+∞
= lim [(1 · ln 1 ’ 1) ’ ( · ln ’ )]
’0+
+ lim [(N · ln N ’ N ) ’ (1 ln 1 ’ 1)]
N ’+∞
= lim [’1 + ] + lim [N ln N ’ N + 1]
’0+ N’+∞
= lim [N ln N ’ N]. This last limit diverges, so
N ’+∞
the integral diverges.
∞ N
dx dx
= lim = lim [arctan x]N
(d)
1 + x 2 N ’+∞ 1 1+x
1
2 N’+∞
1
π π π
= lim (arctan N ’ arctan 1) = ’ = .
N ’+∞ 2 4 4
300 Solutions to Exercises

Chapter 6
2 ln a ’ 3 ln b ’ 4 ln c ’ ln d
1. (a)
ln 2
(b)
ln 3
(c) 3x + 4 ln z ’ 3 ln w
1
(d) 2w +
2
2. We do (a) and (b).
(a)
3x · 5’x = 2x · e3
x ln 3 ’ x ln 5 = x ln 2 + 3
x · [ln 3 ’ ln 5 ’ ln 2] = 3
3
x= .
ln 3 ’ ln 5 ’ ln 2
(b)
3x
= 10x · 102
’x · 42x
5
x log10 3 + x log10 5 ’ 2x log10 4 = x + 2
x[log10 3 + log10 5 ’ 2 log10 4 ’ 1] = 2
2
x= .
log10 3 + log10 5 ’ 2 log10 4 ’ 1
2x · cos(x 2 )
3. (a)
sin(x 2 )
2 1

(b)
x’1
x
sin(ex )
· cos(ex ) · ex
(c) e
1
cos(ln x) ·
(d)
x
’e x ’ e’x · 3x 2 ’ e’x · 6x ’ 6e’x + C
’x 3
4. (a)
x3 2 2 x3
23
ln x ’ x ln x + +C
(b)
3 9 39
e
ln2 x 1
=
(c)
2 2
1
2
ln(e + 1) = ln(e2 + 1) ’ ln(e + 1)
x
(d) 1
301
Chapter 6

5. We do (a) and (b).
x2 + 1
Let A = ·3
x3
(a) . Then
x ’x
ln A = 3 ln x + ln(x 2 + 1) ’ ln(x 3 ’ x)
hence
3x 2 ’ 1
3 2x
dA/dx d
= ln A = + 2 ’3 .
x +1 x ’x
A dx x
Multiplying through by A gives

3 x +1 3x 2 ’ 1
2 3 2x
dA
= x·3 · +2 ’3 .
x ’x x +1 x ’x
dx x
sin x · (x 3 + x)
Let A =
(b) . Then
x 2 (x + 1)
ln A = ln sin x + ln(x 3 + x) ’ ln x 2 ’ ln(x + 1)
hence
3x 2 + 1 2x
cos x 1
dA/dx d
= ln A = +3 ’ 2’ .
x+1
x +x
sin x
A dx x
Multiplying through by A gives
sin x · (x 3 + x) 3x 2 + 1 2x
cos x 1
dA
= · +3 ’ 2’ .
x+1
x 2 (x + 1) x +x
sin x
dx x
6. Let R(t) denote the amount of substance present at time t. Let noon on
January 10 correspond to t = 0 and noon on February 10 correspond to
t = 1. Then R(0) = 5 and R(1) = 3. We know that
R(t) = P · eKt .
Since
5 = R(0) = P · eK·0 ,
we see that P = 5. Since
3 = R(1) = 5 · eK·1 ,
we ¬nd that K = ln 3/5. Thus
t
3
R(t) = 5 · e =5·
t ln(3/5)
.
5
302 Solutions to Exercises

Taking March 10 to be about t = 2, we ¬nd that the amount of radioactive
material present on March 10 is
2
3 9
R(2) = 5 · =.
5 5
7. Let the amount of bacteria present at time t be
B(t) = P · eKt .
Let t = 0 be 10:00 a.m. We know that B(0) = 10000 and B(3) = 15000.
Thus
10000 = B(0) = P · eK·0
so P = 10000. Also
15000 = B(3) = 10000 · eK·3
hence
1
K= · ln(3/2).
3
As a result,
B(t) = 10000 · et·[1/3] ln(3/2)
or
t/3
3
B(t) = 10000 · .
2
We ¬nd that, at 2:00 p.m., the number of bacteria is
4/3
3
B(4) = 10000 · .
2
8. If M(t) is the amount of money in the account at time t then we know that
M(t) = 1000 · e6t/100 .
Here t = 0 corresponds to January 1, 2005. Then, on January 1, 2009, the
amount of money present is
M(4) = 1000 · e6·4/100 ≈ 1271.25.
1
· 1 · ex + x · ex
9. (a)
1 ’ (x · ex )2
1 1
·
(b)
1 + (x/[x + 1])2 (x + 1)2
303
Chapter 7

2x + 1
1
·2
(c)
1 + [ln(x 2 + x)]2 x + x
1
· sec2 x
(d)
| tan x| [tan x]2 ’ 1
Tan’1 x 2 + C
10. (a)
Sin’1 x 3 + C
(b)
π
π/2
’1
= Sin’1 1 ’ Sin’1 0 = .
2
(c) Sin (sin x)
2
0

1 1 2x
dx
= √ · Tan’1 √ +C

(d)
1 + [ 2/5x]2
5 5
10

Chapter 7
1. We do (a), (b), (c), (d).
Let u = log2 x and dv = 1 dx. Then
(a)
1
log2 x dx = log2 x · x ’ x · 2 log x · dx
x
= x log2 x ’ 2 log x dx.

Now let u = log x and dv = 1 dx. Then
1
log2 x dx = x log2 x ’ 2 log x · x ’ x· dx
x
= x log2 x ’ 2x log x + 2x + C.
Let u = x and dv = e3x dx. Then
(b)
e3x e3x
x · e x dx = x · ’ · 1 dx
3
3 3
e3x e3x
=x· ’ + C.
3 9
Let u = x 2 and dv = cos x dx. Then
(c)

x 2 cos x dx = x 2 · sin x ’ sin x · 2x dx.

Now let u = 2x and dv = sin x dx. Then

x 2 cos x dx = x 2 · sin x ’ 2x · (’ cos x) ’ (’ cos x) · 2 dx

= x 2 sin x + 2x cos x ’ 2 sin x + C.
304 Solutions to Exercises

Notice that t sin 3t cos 3t dt = 1 t sin 6t dt. Now let u = t and
(d) 2
dv = sin 6t dt. Then
1 1 1 1
t sin 6t dt = t · ’ cos 6t ’ ’ cos 6t · 1 dt
2 2 6 6
1
t
= ’ cos 6t + sin 6t + C.
12 72
2. We do (a), (b), (c), (d).
’1/7
1 1/7
= +
(a) hence
(x + 2)(x ’ 5) x+2 x’5
’1/7 1/7
dx
= +
(x + 2)(x ’ 5) x+2 x’5
’1 1
= ln |x + 2| + ln |x ’ 5| + C.
7 7
’x/2 + 1/2
1 1/2
= +
(b) hence
x+1
(x + 1)(x 2 + 1) x2 + 1
’x/2
1/2 1/2
dx
= dx + dx + dx
x +1
(x +1)(x 2 +1) x 2 +1 x 2 +1
1 1 1
= ln |x +1|’ ln |x 2 +1|+ Tan’1 x +C.
2 4 2
Now x 3 ’ 2x 2 ’ 5x + 6 = (x ’ 3)(x + 2)(x ’ 1). Then
(c)
’1/6
1 1/10 1/15
= + + .
x’3 x+2 x’1
x 3 ’ 2x 2 ’ 5x + 6
As a result,
’1/6
1/10 1/15
dx
= dx + dx + dx
x ’3 x +2 x ’1
x 3 ’2x 2 ’5x +6
1 1 1
= ln |x ’3|+ ln |x +2|’ ln |x ’1|+C.
10 15 6
Now x 4 ’ 1 = (x 2 ’ 1)(x 2 + 1) = (x ’ 1)(x + 1)(x 2 + 1).
(d)
Hence
’x/2
1/4 1/4
x
= + +2 .
x’1 x+1 x +1
x4 ’ 1
We conclude that
1 1 1
x dx
= ln |x ’ 1| + ln |x + 1| ’ ln |x 2 + 1| + C.
x4 + 1 4 4 4
305
Chapter 7

3. We do (a), (b), (c), (d).
Let u = sin x, du = cos x dx. Then the integral becomes
(a)
(1 + u2 )3
(1 + u ) 2u du = + C.
22
3
Resubstituting x, we obtain the ¬nal answer
(1 + sin2 x)3
(1 + sin x) 2 sin x cos x dx = + C.
2 2
3
√ √
Let u = x, du = 1/[2 x] dx. Then the integral becomes
(b)

2 sin u du = ’2 cos u + C.

Resubstituting x, we obtain the ¬nal answer


sin x
dx = ’2 cos x + C.

x
Let u = ln x, du = [1/x] dx. Then the integral becomes
(c)
1 1
cos u sin u du = sin 2u du = ’ cos 2u + C.
2 4
Resubstituting x, we obtain the ¬nal answer
cos(ln x) sin(ln x) 1
dx = ’ cos(2 ln x) + C.
4
x
Let u = tan x, du = sec2 x dx. Then the integral becomes
(d)

eu du = eu + C.

Resubstituting x, we obtain the ¬nal answer

etan x sec2 x dx = etan x + C.

4. We do (a), (b), (c), (d).
Let u = cos x, du = ’ sin x dx. Then the integral becomes
(a)
u3
’ u du = ’ + C.
2
3
Resubstituting x, we obtain the ¬nal answer
cos3 x
sin x cos x dx = ’ + C.
2
3
306 Solutions to Exercises

(b) Write

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