2x-3(r2-1) (2x+3)

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Solution

The integral is given as follows

I= ∫ ( 2x−3 ) ( x 2 −1 )( 2x+3 ) dx

Use partial fraction rule to simplify the fraction.

( 2x−3 ) ( x+1 )( x−1 )( 2x+3 ) = A ( x+1 ) + B ( x−1 ) + C ( 2x+3 ) 2x−3=A( x−1 )( 2x+3 )+B( x+1 )( 2x+3 )+C( x 2 −1 )

Substitute x=1then,

B= −1 10

Substitute x=−1then,

A= 5 2

Substitute x=0then,

−3A+3B−C=−3

Substitute the values and solve for C

C= −24 5

On Integrating, we get

I= ∫ ( 2x−3 )dx ( x+1 )( x−1 )( 2x+3 ) = 5 2 ∫ dx x+1 − 1 10 ∫ dx x−1 − 24 5 ∫ dx ( 2x+3 ) = 5 2 log| x+1 |− 1 10 log| x−1 |− 24 5 × 1 2 log| 2x+3 |+C I= 5 2 log| x+1 |− 1 10 log| x−1 |− 12 5 log| 2x+3 |+C

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