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

Let $\displaystyle \mathrm{I}=\int x \cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right) d x$I=∫xcos−1(1+x21−x2​)dx $$\therefore \mathrm{I}=2 \int_{\text {II }} x \cdot \tan ^{-1} x d x$$∴I=2∫II ​x⋅tan−1xdx Applying Integration by parts $$\begin{array}{r} \mathrm{I}=2\left[\tan ^{-1} x \int x d x-\int\left(\frac{d}{d x}\left(\tan ^{-1} x\right) \int x d x\right) d x\right] \\ \mathrm{I}=2\left[\frac{x^2}{2} \tan ^{-1} x-\int \frac{1}{1+x^2} \times \frac{x^2}{2} d x\right]+c \\ \mathrm{I}=2\left[\frac{x^2}{2} \tan ^{-1} x-\frac{1}{2} \int \frac{x^2+1-1}{x^2+1} d x\right]+c \\ \mathrm{I}=2\left[\frac{x^2}{2} \tan ^{-1} x-\frac{1}{2} \int \frac{x^2+1}{x^2+1} d x+\frac{1}{2} \int \frac{1}{1+x^2} d x\right]+c \end{array}$$I=2[tan−1x∫xdx−∫(dxd​(tan−1x)∫xdx)dx]I=2[2x2​tan−1x−∫1+x21​×2x2​dx]+cI=2[2x2​tan−1x−21​∫x2+1x2+1−1​dx]+cI=2[2x2​tan−1x−21​∫x2+1x2+1​dx+21​∫1+x21​dx]+c​ $$\begin{aligned} \mathrm{I} &=2\left[\frac{x^2}{2} \tan ^{-1} x-\frac{1}{2} \int 1 . d x+\frac{1}{2} \tan ^{-1} x\right]+c \\ \mathrm{I} &=2\left[\frac{x^2}{2} \tan ^{-1} x-\frac{x}{2}+\frac{1}{2} \tan ^{-1} x\right]+c \\ \mathrm{I} &=x^2 \tan ^{-1} x+\tan ^{-1} x-x+c \\ & \text { or } I=-x+\left(x^2+1\right) \tan ^{-1} x+c \end{aligned}$$III​=2[2x2​tan−1x−21​∫1.dx+21​tan−1x]+c=2[2x2​tan−1x−2x​+21​tan−1x]+c=x2tan−1x+tan−1x−x+c or I=−x+(x2+1)tan−1x+c​