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

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