Solução_Calculo_Stewart_6e

F.
If we now let u = t 2 ,then
SECTIONTX.10
7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS ¤ 317
1 − t 2
(t 2 +3)(t 2 +1) 2 = 1 − u
(u +3)(u +1) 2 = A
u +3 +
B
u +1 +
C
(u +1) 2 ⇒
1 − u = A(u +1) 2 + B(u +3)(u +1)+C(u +3).Setu = −1 to get 2=2C,soC =1.Setu = −3 to get 4=4A,so
A =1.Setu =0to get 1=1+3B +3,soB = −1. So
I =
1
0
8t
t 2 +3 −
8t
t 2 +1 +
8t
(t 2 +1) 2 dt = 4ln(t 2 +3)− 4ln(t 2 +1)− 4 1
t +1 2 0
=(4ln4− 4ln2− 2) − (4 ln 3 − 0 − 4) = 8 ln 2 − 4ln2− 4ln3+2=4ln 2 3 +2
63. By long division,
x 2 +1 3x +1
= −1+
3x − x2 3x − x .Now 2
3x +1 3x +1
=
3x − x2 x(3 − x) = A x + B
3 − x ⇒ 3x +1=A(3 − x)+Bx. Setx =3to get 10 = 3B,soB = 10 .Setx =0to
3
get 1=3A,soA = 1 3
. Thus, the area is
2
1
x 2
+1
2
1 10
3x − x dx = 3
−1+
2 x + 3
1
3 − x
dx = −x + 1 10
ln |x| − ln |3 − x| 2
3 3 1
= −2+ 1 3 ln 2 − 0 − −1+0− 10 3 ln 2 = −1+ 11 3 ln 2
65.
P + S
P [(r − 1)P − S] = A P + B
(r − 1)P − S
⇒ P + S = A [(r − 1)P − S]+BP =[(r − 1)A + B] P − AS ⇒
(r − 1)A + B =1, −A =1 ⇒ A = −1, B = r. Now
t =
so t = − ln P +
P + S
−1
P [(r − 1)P − S] dP = P + r
dP
dP = −
(r − 1)P − S
P + r
r − 1
r ln|(r − 1)P − S| + C. Herer =0.10 and S =900,so
r − 1
r − 1
(r − 1)P − S dP
t = − ln P + 0.1 ln|−0.9P − 900| + C = − ln P − 1 ln(|−1||0.9P + 900|) =− ln P − 1 ln(0.9P + 900) + C.
−0.9 9 9
When t =0, P =10,000,so0=− ln 10,000 − 1 ln(9900) + C. Thus,C =ln10,000 + 1 9 9
ln 9900 [≈ 10.2326], so our
equation becomes
10,000
ln(0.9P + 900) = ln
P
+ 1 9 ln 1100
0.1P + 100 =ln10,000 + 1 11,000
ln
P 9 P +1000
t =ln10,000 − ln P + 1 9 ln 9900 − 1 9
=ln 10,000
P
67. (a) In Maple, we define f(x),andthenuseconvert(f,parfrac,x); to obtain
f(x) = 24,110/4879
5x +2
+ 1 9 ln 9900
0.9P +900
− 668/323
2x +1 − 9438/80,155 (22,098x +48,935)/260,015
+
3x − 7
x 2 + x +5
In Mathematica, we use the command Apart,andinDerive,weuseExpand.