We will use integration by substitution, as well as the integrals
∫
1
x
d
x
=
ln
|
x
|
+
C
and
∫
1
d
x
=
x
+
C
∫
x
3
x
2
+
1
d
x
=
∫
x
2
x
2
+
1
x
d
x
=
1
2
∫
(
x
2
+
1
)
−
1
x
2
+
1
2
x
d
x
Let
u
=
x
2
+
1
⇒
d
u
=
2
x
d
x
. Then
1
2
∫
(
x
2
+
1
)
−
1
x
2
+
1
2
x
d
x
=
1
2
∫
u
−
1
u
d
u
=
1
2
∫
(
1
−
1
u
)
d
u
=
1
2
(
u
−
ln
|
u
|
)
+
C
=
x
2
+
1
2
−
ln
(
x
2
+
1
)
2
+
C
=
x
2
2
−
ln
(
x
2
+
1
)
2
+
1
2
+
C
=
x
2
−
ln
(
x
2
+
1
)
2
+
C
Final answer
