Solução_Calculo_Stewart_6e

F.
SECTIONTX.10
7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS ¤ 313
so B = − 1 3 , C = − 2 3
⇒
1
1
x 3 − 1 dx = 3
−
1
x − 1 dx + x − 2 3 3
x 2 + x +1 dx = 1 ln |x − 1| − 1
3
3
= 1 ln |x − 1| − 1
x +1/2
3
3 x 2 + x +1 dx − 1
3
(3/2) dx
(x +1/2) 2 +3/4
x +2
x 2 + x +1 dx
= 1 3 ln |x − 1| − 1 6 ln x 2 + x +1 − 1 2
2√3
tan −1
x +
1
2
√
3
2
+ K
= 1 ln |x − 1| − 1 3 6 ln(x2 + x +1)− √ 1
3
tan −1 √
1
3
(2x +1) + K
33. Let u = x 4 +4x 2 +3, so that du =(4x 3 +8x) dx =4(x 3 +2x) dx, x =0 ⇒ u =3,andx =1 ⇒ u =8.
35.
37.
1
Then
0
x 3
+2x
8
x 4 +4x 2 +3 dx =
3
1 1
u 4 du = 1 8
ln |u|
4 = 1 3
4 (ln 8 − ln 3) = 1 4 ln 8 3 .
1
x(x 2 +4) 2 = A x + Bx + C
x 2 +4 + Dx + E
(x 2 +4) 2 ⇒ 1=A(x 2 +4) 2 +(Bx + C)x(x 2 +4)+(Dx + E)x. Setting x =0
gives 1=16A,soA = 1 . Now compare coefficients.
16
1= 1
16 (x4 +8x 2 +16)+(Bx 2 + Cx)(x 2 +4)+Dx 2 + Ex
1= 1
16 x4 + 1 2 x2 +1+Bx 4 + Cx 3 +4Bx 2 +4Cx + Dx 2 + Ex
1= 1
16 + B x 4 + Cx 3 + 1
2 +4B + D x 2 +(4C + E)x +1
So B + 1 =0 ⇒ B = − 1 , C =0, 1 +4B + D =0 ⇒ D = − 1 16 16 2 4
,and4C + E =0 ⇒ E =0.Thus,
dx
1
x(x 2 +4) = 16
2 x + − 1 x 16
x 2 +4 + − 1 x
4
dx = 1 (x 2 +4) 2 16 ln |x| − 1 16 · 1
2 ln x 2 +4
1 − − 1 1
4 2 x 2 +4 + C
= 1 16
1
1
ln |x| −
32 ln(x2 +4)+
8(x 2 +4) + C
x 2 − 3x +7
(x 2 − 4x +6) = Ax + B
2 x 2 − 4x +6 + Cx + D
⇒ x 2 − 3x +7=(Ax + B)(x 2 − 4x +6)+Cx + D ⇒
(x 2 − 4x +6) 2
x 2 − 3x +7=Ax 3 +(−4A + B)x 2 +(6A − 4B + C)x +(6B + D). SoA =0, −4A + B =1 ⇒ B =1,
6A − 4B + C = −3 ⇒ C =1, 6B + D =7 ⇒ D =1.Thus,
I =
=
x 2
− 3x +7
(x 2 − 4x +6) dx = 1
2 x 2 − 4x +6 + x +1
dx
(x 2 − 4x +6) 2
1
(x − 2) 2 +2 dx + x − 2
(x 2 − 4x +6) dx + 3
2 (x 2 − 4x +6) dx 2
= I 1 + I 2 + I 3.
[continued]