Solução_Calculo_Stewart_6e

F.
314 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION
I 1 =
1
(x − 2) 2 + √ 2 2 dx = √ 1
tan −1 x − 2
√ + C 1
2 2
I 2 = 1
2x − 4
2 (x 2 − 4x +6) dx = 1 1 2 2 u du = 1 −
2
1
1
+ C 2 2 = −
u
2(x 2 − 4x +6) + C 2
I 3 =3
= 3 √ 2
4
1
(x − 2) 2 + √ 2 2 2 dx =3
sec 2 θ
sec 4 θ dθ = 3 √
2
4
1
[2(tan 2 θ +1)] 2 √
2sec 2 θdθ
cos 2 θdθ= 3 √ 2
4
1
(1 + cos 2θ) dθ
2
= 3 √ 2 θ +
1
2
8
sin 2θ + C 3 = 3 √
2 x − 2
tan −1 √ + 3 √ 2
8
2 8
= 3 √ 2
8
= 3 √ 2
8
x − 2
tan −1 √ + 3 √ 2
·
2 8
x − 2
√
x2 − 4x +6 ·
x − 2
tan −1 3(x − 2)
√ +
2 4(x 2 − 4x +6) + C3
So I = I 1 + I 2 + I 3 [C = C 1 + C 2 + C 3]
= 1 √
2
tan −1 x − 2
√
2
+
√
4 2
=
8
+ 3 √
2 x − 2
tan −1 √ +
8
2
TX.10
√
2
√
x2 − 4x +6 + C 3
−1
2(x 2 − 4x +6) + 3 √
2 x − 2
tan −1 √ +
8
2
x − 2= √ 2tanθ,
dx = √ 2sec 2 θdθ
1
2 · 2sinθ cos θ + C 3
3(x − 2)
4(x 2 − 4x +6) + C
3(x − 2) − 2
4(x 2 − 4x +6) + C = 7 √
2 x − 2
tan −1 √ +
8
2
3x − 8
4(x 2 − 4x +6) + C
39. Let u = √ x +1.Thenx = u 2 − 1, dx =2udu ⇒
√
dx
x √ x +1 = 2udu
(u 2 − 1) u =2 du
u 2 − 1 =ln u − 1
x +1− 1
u +1 + C =ln √ + C.
x +1+1
41. Let u = √ x,sou 2 = x and dx =2udu.Thus,
Multiply
16
9
√ x
x − 4 dx = 4
3
=2+8
u
4
u 2 − 4 2udu=2
4
3
du
(u +2)(u − 2) ()
3
u 2 4
u 2 − 4 du =2
3
1+ 4
du
u 2 − 4
[by long division]
1
(u +2)(u − 2) = A
u +2 + B by (u +2)(u − 2) to get 1=A(u − 2) + B(u +2). Equating coefficients we
u − 2
get A + B =0and −2A +2B =1. Solving gives us B = 1 and A = − 1 ,so 1
4 4
(u +2)(u − 2) = −1/4
u +2 + 1/4 and ()is
u − 2
2+8
4
3
−1/4
u +2 + 1/4
4
4
du =2+8 − 1
u − 2
ln |u +2| + 1 ln |u − 2| =2+ 2ln|u − 2| − 2ln|u +2|
4 4
3 3
=2+2
ln
u − 2
u +2
4
3
=2+2 ln 2 6 − ln 1 5
=2+2ln 5 or 2+ln
5 2
=2+ln 25
3 3
9
=2+2ln
2/6
1/5