Question

1x(d+1) equals(A) log kl-2log +D+C(C) -log lxl+^log +)C(D)23.(B) logll+ C^logll+ log (a2+1)+c

Answer 1

The integral is given as,

I= ∫ dx x( x 2 +1 )

Use partial fraction rule to simplify fraction.

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

Substitute x=0then,

A=1

Equate the coefficients of x 2 ,xand constant term.

A+B=0 C=0

So, B=−1

Substitute the values and integrate.

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

Thus, the correct option is (A).