Integration 229(1−x)^5 /^2 − 10516 x(1−x)^7 /^2 and simplify to
−2
105
(1−x)^3 /^2[
15 x^2 + 12 x+ 8]
+C.- For
∫
3 x^2 sinxdx, use integration by
parts withu= 3 x^2 ,du= 6 xdx,
dv∫ =sinxdx, andv=−cosx.
3 x^2 sinxdx=− 3 x^2 cosx+
[
6 xsinx−∫
6 sinxdx]
=− 3 x^2 cosx+6 xsinx+6 cosx+C.- Factor the denominator so that∫
xdx
x^2 − 3 x− 4
=
∫
xdx
(x− 4 )(x+ 1 ).
Use a partial fraction decomposition,
x
(x− 4 )(x+ 1 )=
A
(x−4)+
B
(x+1)
, which
impliesAx+A+Bx− 4 B=x. Solve
A+B=1 andA− 4 Bto find
A=4
5
andB=1
5
. Integrate
∫
xdx
x^2 − 3 x− 4
=
∫
4 / 5
x− 4
dx+
∫
1 / 5
x+ 1
dx=4
5
ln∣∣
x− 4∣∣+
1
5
ln∣∣
x+ 1∣∣
+C=
1
5
ln∣∣
∣(x−^4 )^4 (x+^1 )∣∣
∣+C.- Factor
∫
dx
x^2 +x=
∫
dx
x(x+ 1 )
and usepartial fractions. If1
x(x+ 1 )
=A
x+
B
(x+1)
,Ax+A+Bx=1 and∫A=−B=^1.
dx
x^2 +x=
∫
dx
x+
∫
−dx
x+ 1
=ln|x|−ln∣∣
x+ 1∣∣
+C=ln∣∣
∣∣ x
x+ 1∣∣
∣∣+C.- Begin with integration by parts, using
u=lnx,du=
dx
x
,dv=
dx
(x+5)^2
,
andv=− 1
x+ 5.
Then∫
lnx
(x+5)^2
dx=
−ln|x|
x+ 5+
∫
dx
x(x+5).
Use partial fractions to decompose
1
x(x+ 5 )=
A
x+
B
(x+5). Solve to find
A=−B=1
5
. Then
∫
lnx
(x+5)^2
dx=
−ln|x|
x+ 5+
1
5
ln|x|−1
5
ln∣∣
x+ 5∣∣=
−ln|x|
x+ 5+
1
5
ln∣∣
∣∣ x
x+ 5∣∣
∣∣+C.10.8 Solutions to Cumulative Review Problems
- (a)Att=4, speed is 5, which is the
greatest on 0≤t≤10.
(b)The particle is moving to the right
when 6<t<10.
27. V=
4
3
πr^3 ;
dV
dt=
(
4
3)
(3)πr^2
dr
dt
= 4 πr^2
dr
dtIfdV
dt= 100
dr
dt
, then 100dr
dt
= 4 πr^2dr
dt⇒ 100.
= 4 πr^2 orr=±√
25
π=±
5
√
π.
Sincer≥0,r=5
√
π
meters.