Question

13. (1 —x) (1+x2)

Answer 1

The integral is given below as,

I= ∫ 2dx ( 1−x )( 1+ x 2 )

Use rule of partial fraction.

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

By equating the coefficients of x 2 ,xand constant term, we get

A−B=0 B−C=0 A+C=2

By solving above equations, we get

A=B=C=1

By substituting the values in integral, we get

I= ∫ 2dx ( 1−x )( 1+ x 2 ) = ∫ dx ( 1−x ) + ∫ ( x+1 )dx ( x 2 +1 ) =− ∫ dx ( x−1 ) + ∫ xdx ( x 2 +1 ) + ∫ dx 1+ x 2

On integrating, we get

I=−log| x−1 |+ 1 2 ∫ 2xdx ( x 2 +1 ) + tan −1 x+C =−log| x−1 |+ 1 2 log| 1+ x 2 |+ tan −1 x+C =log| 1 x−1 |+ 1 2 log| 1+ x 2 |+ tan −1 x+C