Solução_Calculo_Stewart_6e

F.
43. Let u = 3√ x 2 +1.Thenx 2 = u 3 − 1, 2xdx=3u 2 du ⇒
x 3
dx (u 3
3√
x2 +1 = − 1) 3 2 u2 du
= 3
(u 4 − u) du
u 2
SECTIONTX.10
7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS ¤ 315
= 3 10 u5 − 3 4 u2 + C = 3 10 (x2 +1) 5/3 − 3 4 (x2 +1) 2/3 + C
45. If we were to substitute u = √ x, then the square root would disappear but a cube root would remain. On the other hand, the
substitution u = 3√ x would eliminate the cube root but leave a square root. We can eliminate both roots by means of the
substitution u = 6√ x. (Note that 6 is the least common multiple of 2 and 3.)
Let u = 6√ x.Thenx = u 6 ,sodx =6u 5 du and √ 3√
x = u 3 , x = u 2 .Thus,
dx 6u 5
√ √ 3
x − x = du
u 3 − u =6 u 5
2 u 2 (u − 1) du =6 u 3
u − 1 du
=6 u 2 + u +1+ 1
du [bylongdivision]
u − 1
=6 1
3 u3 + 1 2 u2 + u +ln|u − 1| + C =2 √ x +3 3√ x +6 6√ x +6ln 6√ x − 1 + C
47. Let u = e x .Thenx =lnu, dx = du u
⇒
e 2x
dx
e 2x +3e x +2 =
u 2
(du/u)
u 2 +3u +2 =
udu
−1
(u +1)(u +2) = u +1 + 2
du
u +2
49. Let u =tant, sothatdu =sec 2 tdt.Then
Now
1
(u +1)(u +2) = A
u +1 + B
u +2
=2ln|u +2| − ln |u +1| + C =ln (ex +2) 2
e x +1
+ C
sec 2
t
tan 2 t +3tant +2 dt =
⇒ 1=A(u +2)+B(u +1).
1
u 2 +3u +2 du =
1
(u +1)(u +2) du.
Setting u = −2 gives 1=−B, soB = −1. Setting u = −1 gives 1=A.
1
1
Thus,
(u +1)(u +2) du = u +1 − 1
du =ln|u +1| −ln |u +2|+C =ln|tan t +1|−ln |tan t +2|+C.
u +2
51. Let u =ln(x 2 − x +2), dv = dx. Thendu = 2x − 1 dx, v = x, and (by integration by parts)
x 2 − x +2
ln(x 2 − x +2)dx = x ln(x 2 2x 2
− x
− x +2)−
x 2 − x +2 dx = x ln(x2 − x +2)− 2+ x − 4
dx
x 2 − x +2
1
= x ln(x 2 (2x − 1)
2
− x +2)− 2x −
x 2 − x +2 dx + 7
dx
2 (x − 1 2 )2 + 7 4
= x ln(x 2 − x +2)− 2x − 1 2 ln(x2 − x +2)+ 7
√
⎡
7
du 2 ⎣
7
2
4 (u2 +1)
=(x − 1 2 )ln(x2 − x +2)− 2x + √ 7tan −1 u + C
=(x − 1 2 )ln(x2 − x +2)− 2x + √ 7tan −1 2x − 1
√ + C
7
where x − 1 2 = √ 7
2 u,
dx = √ 7
2 du,
(x − 1 2 )2 + 7 4 = 7 4 (u2 +1)
⎤
⎦