Answer :
Let,
3
x
2
+
2
x
+
1
=
t
=>
(
6
x
+
2
)
d
x
=
d
t
=>
2
(
3
x
+
1
)
d
x
=
d
t
So, given equation becomes,
=>
∫
d
t
2
t
3
=>
−
1
8
t
4
+
C
=>
−
1
8
(
3
x
2
+
2
x
+
1
)
4
+
C
Let,
3
x
2
+
2
x
+
1
=
t
=>
(
6
x
+
2
)
d
x
=
d
t
=>
2
(
3
x
+
1
)
d
x
=
d
t
So, given equation becomes,
=>
∫
d
t
2
t
3
=>
−
1
8
t
4
+
C
=>
−
1
8
(
3
x
2
+
2
x
+
1
)
4
+
C