Answers and Explanations 521
Answers
801–900
878.
3
6
3
3
18
3 3
108
1
108
6
2
6
6
(ln )
ln
−
+
−
Here’s the given integral:
x
x dx
5
2
1
3 (ln )
∫
Begin by using integration by parts with u = (ln x)
2
so that du
x x
dx
= 2
1
(ln )
and
let dv = x
5 
dx so that v x
=
6
6
. This gives you the following:
x
x dx x
x
x
x x
dx
x
x
5
1
3
2
6
2
1
3
6
1
3
6
2
1
3
6
6
2
1
6
1
(ln )
(ln )
(ln )
(ln )
∫
∫
=
−
(
)
=
− 3 3
5
1
3 x
xdx
ln
∫
To evaluate the new integral, use integration by parts with u 1 = ln x so that du
x
dx
1
1
=
and let dv 1 = x
5 
dx so that v
x
1
6
6
=
. This gives you the following:
x
x
x
x dx
x
x
x
x
x
x
dx
6
2
1
3
5
1
3
6
2
1
3
6
1
3
6
1
3
6
1
3
6
1
3 6
6
1
(ln )
ln
(ln )
ln
−
=
−
−
∫
∫
 

 


 
=
−
+
=
−
∫
x
x
x
x
xdx
x
x
x
6
2
1
3
6
1
3
5
1
3
6
2
6
6
1 8
1
3
1
6
6
1 8
(ln )
ln
(ln )
l ln
(ln )
ln
x
x
+






=
−
+
− − +
(
)
=
1
108
3
6
3
3
18
3 3
108
0 0 1
108
3
6
6
1
3
6
2
6
6
6 ( (ln )
ln
3
3
18
3 3
108
1
108
2
6
6
−
+
−
879.
1
3
2
2
e
x e
x
C
x
x
cos( )
sin( )
+
)
(
+
The given integral is
e
x dx
x cos( )
2
∫
Begin by using integration by parts with u = cos(2x) so that du = –2sin(2x)dx and
let dv = e
x 
dx so that v = e
x
. Using the integration by parts formula gives you the
following:
e
x dx
e
x
e
x dx
e
x
e
x
x
x
x
x
cos( )
cos( )
s in( )
cos( )
s in(
2
2
2
2
2
2
2
∫
∫
=
−
−
(
)
=
+
x x dx
)
∫