Solução_Calculo_Stewart_6e

F.
310 ¤ CHAPTER 7 TECHNIQUES OF INTEGRATION
TX.10
7.4 Integration of Rational Functions by Partial Fractions
1. (a)
(b)
3. (a)
(b)
5. (a)
(b)
7.
9.
11.
2x
(x + 3)(3x +1) = A
x +3 + B
3x +1
1
x 3 +2x 2 + x = 1
x(x 2 +2x +1) = 1
x(x +1) 2 = A x +
x 4 +1
x 5 +4x = x4 +1
3 x 3 (x 2 +4) = A x + B x + C 2 x + Dx + E
3 x 2 +4
B
x +1 +
1
(x 2 − 9) 2 = 1
[(x +3)(x − 3)] 2 = 1
(x +3) 2 (x − 3) 2 = A
x +3 +
x 4
x 4 − 1 = (x4 − 1) + 1
=1+ 1
x 4 − 1 x 4 − 1
=1+
C
(x +1) 2
[or use long division] =1+
1
(x − 1)(x +1)(x 2 +1) =1+ A
x − 1 + B
x +1 + Cx + D
x 2 +1
t 4 + t 2 +1
(t 2 +1)(t 2 +4) 2 = At + B
t 2 +1 + Ct + D
t 2 +4 + Et + F
(t 2 +4) 2
B
(x +3) 2 + C
x − 3 +
x (x − 6) + 6
x − 6 dx = dx = 1+ 6
dx = x +6ln|x − 6| + C
x − 6
x − 6
1
(x 2 − 1)(x 2 +1)
D
(x − 3) 2
x − 9
(x +5)(x − 2) = A
x +5 + B . Multiply both sides by (x +5)(x − 2) to get x − 9=A(x − 2) + B(x +5)(∗),or
x − 2
equivalently, x − 9=(A + B)x − 2A +5B. Equating coefficients of x on each side of the equation gives us 1=A + B (1)
and equating constants gives us −9 =−2A +5B (2). Adding two times (1)to(2)givesus−7 =7B ⇔ B = −1 and
hence, A =2. [Alternatively, to find the coefficients A and B, we may use substitution as follows: substitute 2 for x in (∗) to
get −7 =7B ⇔ B = −1, then substitute −5 for x in (∗) to get −14 = −7A ⇔ A =2.] Thus,
x − 9
2
(x +5)(x − 2) dx = x +5 + −1
dx =2ln|x +5| − ln |x − 2| + C.
x − 2
1
x 2 − 1 = 1
(x +1)(x − 1) = A
x +1 + B . Multiply both sides by (x +1)(x − 1) to get 1=A(x − 1) + B(x +1).
x − 1
Substituting 1 for x gives 1=2B ⇔ B = 1 2 . Substituting −1 for x gives 1=−2A ⇔ A = − 1 2 . Thus,
3
2
13.
1
3
x 2 − 1 dx =
ax
x 2 − bx dx =
2
−1/2
x +1 + 1/2
dx = − 1 2
x − 1
ln |x +1| + 1 ln |x − 1| 3
2 2
= − 1 2 ln 4 + 1 2 ln 2 − − 1 2 ln 3 + 1 2 ln 1 = 1 2 (ln2+ln3− ln 4) or
1
2 ln 3 2
ax
x(x − b) dx =
a
dx = a ln |x − b| + C
x − b