Solução_Calculo_Stewart_6e

F.
SECTIONTX.10
7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS ¤ 311
15.
x 3 − 2x 2 − 4
x 3 − 2x 2 =1+
−4
x 2 (x − 2) .Write
−4
x 2 (x − 2) = A x + B x 2 +
C
x − 2 . Multiplying both sides by x2 (x − 2) gives
−4 =Ax(x − 2) + B(x − 2) + Cx 2 . Substituting 0 for x gives −4 =−2B ⇔ B =2. Substituting 2 for x gives
−4 =4C ⇔ C = −1. Equating coefficients of x 2 ,weget0=A + C,soA =1.Thus,
4
3
x 3 − 2x 2 − 4
x 3 − 2x 2 dx =
4
3
1+ 1 x + 2 x − 1
dx = x +ln|x| − 2 4 2 x − 2
x − ln |x − 2| 3
= 4+ln4− 1 2 − ln 2 − 3+ln3− 2 3 − 0 = 7 6 +ln2 3
17.
4y 2 − 7y − 12
y(y +2)(y − 3) = A y +
B
y +2 +
C
y − 3
⇒
4y 2 − 7y − 12 = A(y +2)(y − 3) + By(y − 3) + Cy(y +2).Setting
y =0gives −12 = −6A,soA =2.Settingy = −2 gives 18 = 10B,soB = 9 5 .Settingy =3gives 3=15C,soC = 1 5 .
Now
2
1
4y 2
− 7y − 12
2
y(y +2)(y − 3) dy =
1
2
y + 9/5
y +2 + 1/5
dy = 2ln|y| + 9
y − 3
ln |y +2| + 1 ln |y − 3| 2
5 5 1
=2ln2+ 9 5 ln 4 + 1 5 ln 1 − 2ln1− 9 5 ln 3 − 1 5 ln 2
=2ln2+ 18 ln 2 − 1 ln 2 − 9 27
ln 3 = ln 2 − 9 ln 3 = 9 (3 ln 2 − ln 3) = 9 ln 8 5 5 5 5 5 5 5 3
19.
1
(x +5) 2 (x − 1) = A
x +5 + B
(x +5) 2 + C
x − 1
⇒ 1=A(x +5)(x − 1) + B(x − 1) + C(x +5) 2 .
Setting x = −5 gives 1=−6B, soB = − 1 .Settingx =1gives 1=36C, soC = 1 6 36
.Settingx = −2 gives
1=A(3)(−3) + B(−3) + C 3 2 = −9A − 3B +9C = −9A + 1 + 1 = −9A + 3 ,so9A = − 1 and A = − 1 .Now
2 4 4 4 36
1
−1/36
(x +5) 2 (x − 1) dx = x +5 − 1/6
(x +5) 2 + 1/36
dx = − 1
x − 1 36 ln |x +5| + 1
6(x +5) + 1 ln |x − 1| + C.
36
21. x
x 2 +4 x 3 + 0x 2 + 0x + 4
x 3 + 4x
−4x +4
x 3 +4
x 2 +4 dx =
x +
By long division, x3 +4 −4x +4
= x +
x 2 +4 x 2 +4 .Thus,
−4x +4
dx = x −
x 2 +4
= 1 2 x2 − 4 · 1
2 ln x 2 +4 +4· 1
2 tan−1 x
2
4x
x 2 +4 + 4
dx
x 2 +2 2
+ C = 1 2 x2 − 2ln(x 2 +4)+2tan −1 x
2
+ C
23.
5x 2 +3x − 2
= 5x2 +3x − 2
= A x 3 +2x 2 x 2 (x +2) x + B x + C
2 x +2 . Multiply by x2 (x +2)to
get 5x 2 +3x − 2=Ax(x +2)+B(x +2)+Cx 2 .Setx = −2 to get C =3,andtake
x =0to get B = −1. Equating the coefficients of x 2 gives 5=A + C ⇒ A =2.So
5x 2
+3x − 2 2
dx =
x 3 +2x 2 x − 1 x + 3
dx =2ln|x| + 1 +3ln|x +2| + C.
2 x +2
x