The integral x^-1(1-x^21+x^2)dx (x0) is equal to:

Question

The integral $\displaystyle\int x\cos^{-1}\left(\dfrac{1-x^2}{1+x^2}\right)dx$ $(x>0)$ is equal to:

Accepted Answer

Since $\cos^{-1}\left(\dfrac{1-x^2}{1+x^2}ight) = 2	an^{-1}x$ for $x>0$:

$I = 2\int x	an^{-1}x\,dx$

By parts: $I = 2\left[	an^{-1}x\cdot\dfrac{x^2}{2} - \int \dfrac{1}{1+x^2}\cdot\dfrac{x^2}{2}\,dxight]$

$= x^2	an^{-1}x - \int \dfrac{x^2}{1+x^2}\,dx = x^2	an^{-1}x - \int\left(1-\dfrac{1}{1+x^2}ight)dx$

$= x^2	an^{-1}x - x + 	an^{-1}x + c = -x+(1+x^2)	an^{-1}x+c$

Solutions · 1