The integral x ⁻¹ ( 1-x^2 1+x^2 ) d x(x>0) is equal to:

Step-by-step solution

Let I = x ⁻¹ ( 1-x^2 1+x^2 ) d x I =2 _ II x ⁻¹ x d x Applying Integration by parts array r I =2 [ ⁻¹ x x d x- ( d d x ( ⁻¹ x ) x d x ) d x ] I =2 [ x^2 2 ⁻¹ x- 1 1+x^2 x^2 2 d x ]+c I =2 [ x^2 2 ⁻¹ x- 1 2 x^2+1-1 x^2+1 d x ]+c I =2 [ x^2 2 ⁻¹ x- 1 2 x^2+1 x^2+1 d x+ 1 2 1 1+x^2 d x ]+c array aligned I &=2 [ x^2 2 ⁻¹ x- 1 2 1 . d x+ 1 2 ⁻¹ x ]+c I &=2 [ x^2 2 ⁻¹ x- x 2 + 1 2 ⁻¹ x ]+c I &=x^2 ⁻¹ x+ ⁻¹ x-x+c & or I=-x+ (x^2+1 ) ⁻¹ x+c aligned