Solução_Calculo_Stewart_6e

F.
340 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION
TX.10
55.
∞
0
∞
0
dx
1
√ =
x (1 + x) 0
dx
√ =
x (1 + x)
dx
∞
√ +
x (1 + x) 1
2udu
u(1 + u 2 )
u = √ x, x = u 2 ,
dx =2udu
dx
1
√ = lim
x (1 + x) t→0 + t
=2
dx
t
√ +lim
x (1 + x) t→∞
1
dx
√ x (1 + x)
.Now
du
1+u 2 =2tan−1 u + C =2tan −1√ x + C,so
dx
√ = lim 2tan
−1 √ x 1
+lim
x (1 + x) t→0 + t 2tan
−1 √ x t
t→∞
1
= lim 2 π
√
t→0 + 4 − 2tan
−1
t
+ lim
√ 2tan
−1
t − 2
π
t→∞ 4 =
π
− 0+2
π
2 2 −
π
= π. 2
1
dx
1
57. If p =1,then
0 x = lim dx
p t→0 + t x = lim [ln x]1 t→0 + t = ∞.
Divergent.
1
dx
1
If p 6= 1,then
0 x = lim dx
p t→0 + t x p
x
−p+1 1
1
= lim
= lim
t→0 + −p +1
t
t→0 + 1 − p
[note that the integral is not improper if p1,thenp − 1 > 0,so
t →∞as t → p−1 0+ , and the integral diverges.
1
1
If p−1 and diverges otherwise.
(p +1)
2
1/t
lim
t→0 + −(p +1)t −(p+2)