Part II: The Answers
516
Answers
801–900
868.
x
x
x
C
tan
l n
−
−
+
+
1
2
1
2
1
The given integral is
tan
−
∫
1 x dx
Use integration by parts with u = tan
−1 
x so that du
x
dx
= +
1
1
2
and let dv = dx so that v = x.
Using the integration by parts formula gives you
tan
t an
−
−
∫
∫
= (
) − +
1
1
2
1
x dx
x x
x
x
dx
To evaluate
x
x
dx
1
2
+
∫
, use the substitution u 1 = 1 + x
2
so that du 1 = 2x
 
dx, or 1
2
1
du x dx
=
.
With these substitutions, you get the following:
tan
t an
tan
l n
t
−
−
−
(
) − +
= (
) −
=
−
+
=
∫
∫
1
2
1
1
1
1
1
1
1
2
1
1
2
x x
x
x
dx
x x
u
du
x
x
u C
x a an
ln
−
−
+
+
1
2
1
2
1
x
x
C
869.
−
+
1
4
4 3
4
ln
The given integral is
ln x
x
dx
2
1
4
∫
Use integration by parts with u = ln x so that du x
dx
= 1
and let dv = x
−2 
dx so that
v
x
x
= −
= −
−1
1 . Using the integration by parts formula gives you the following:
x
xdx
x x
x x
dx
x
x
x
−
−
∫
∫
=
−
( )





 −
− ⋅
( )
= −
(
) + (
2
1
4
1
4
1
4
1
4
2
1
1 1
1
ln
ln
ln
) )
= −
−
(
)
= −
−
(
) − − −
(
)






= −
∫ 1
4
1
4
1
1
1
4
4 1
4
1
1
1 1
1
1
4
4
dx
x
x x
ln
ln
ln
ln + + 3
4