Page 317 - Calculus Demystified

304

Solutions to Exercises
(d) Notice that t sin 3t cos 3tdt = 1 t sin 6tdt. Now let u = t and
2
dv = sin 6tdt. Then

1 1 1 1
t sin 6tdt = t · − cos 6t − − cos 6t · 1 dt
2 2 6 6
t 1
=− cos 6t + sin 6t + C.
12 72
2. We do (a), (b), (c), (d).
1 −1/7 1/7
(a) = + hence
(x + 2)(x − 5) x + 2 x − 5

dx −1/7 1/7
= +
(x + 2)(x − 5) x + 2 x − 5
−1 1
= ln |x + 2|+ ln |x − 5|+ C.
7 7
1 1/2 −x/2 + 1/2
(b) = + hence
2
2
(x + 1)(x + 1) x + 1 x + 1

dx 1/2 −x/2 1/2
= dx+ dx+ dx
2
2
2
(x+1)(x +1) x+1 x +1 x +1
1 1 2 1 −1
= ln|x+1|− ln|x +1|+ Tan x+C.
2 4 2
2
3
(c) Now x − 2x − 5x + 6 = (x − 3)(x + 2)(x − 1). Then
1 1/10 1/15 −1/6
= + + .
2
3
x − 2x − 5x + 6 x − 3 x + 2 x − 1
As a result,

dx 1/10 1/15 −1/6
= dx+ dx+ dx
2
3
x −2x −5x+6 x−3 x+2 x−1
1 1 1
= ln|x−3|+ ln|x+2|− ln|x−1|+C.
10 15 6
4
2
2
2
(d) Now x − 1 = (x − 1)(x + 1) = (x − 1)(x + 1)(x + 1).
Hence
x 1/4 1/4 −x/2
= + + .
2
4
x − 1 x − 1 x + 1 x + 1
We conclude that

xdx 1 1 1 2
= ln |x − 1|+ ln |x + 1|− ln |x + 1|+ C.
4
x + 1 4 4 4