Answers and Explanations 517
Answers
801–900
870.
4e
3
Here’s the given integral:
e dx
x
1
9
∫
Begin by using the substitution w
x
=
so that w
2
= x and 2w
 
dw = dx. You can find the
new limits of integration by noting that if x = 9, then w =
=
9 3, and if x = 1, then
w =
=
1 1. These substitutions give you the following:
2
1
3 we dw
w
∫
Use integration by parts with u = 2w so that du = 2dw and let dv = e
w 
dw so that v = e
w
.
Using the integration by parts formula gives you the following:
2
2
2
2
2
6
2
2
2
1
3
1
3
1
3
1
3
3
3
1
1
we dw
we
e dw
we
e
e
e
e
e
w
w
w
w
w
∫
∫
=
−
=
−
(
)
=
−
(
) − −
(
)
= = 4
3
e
871.
486
5
9 968
25
ln −
The given integral is
x
xdx
3 2
1
9
ln
∫
Use integration by parts with u = ln x so that du x
dx
= 1
and let dv = x
3/2 
dx so that
v
x
= 2
5
5 2 to get the following:
x
xdx
x
x
x
x
dx
x
x
x d
3 2
1
9
5 2
1
9
5 2
1
9
5 2
1
9
3 2
1
9
2
5
2
5
1
2
5
2
5
ln
ln
ln
∫
∫
∫
=
−
( )
=
−
x x
x
x
x
=
− ( )






=
−
(
) −
−
2
5
2
5
2
5
2
5
9
9 4
25
9
2
5
1
1 4
5 2
5 2
1
9
5 2
5 2
5 2
ln
ln
ln 2 25
1
486
5
9 972
25
4
25
486
5
9 968
25
5 2
(
)
=
−
+
=
−
ln
ln