Let $\mathrm{I}=\int x \cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right) d x$
$$
\therefore \mathrm{I}=2 \int_{\text {II }} x \cdot \tan ^{-1} x d x
$$
Applying Integration by parts
$$
\begin{array}{r}
\mathrm{I}=2\left[\tan ^{-1} x \int x d x-\int\left(\frac{d}{d x}\left(\tan ^{-1} x\right) \int x d x\right) d x\right] \\
\mathrm{I}=2\left[\frac{x^2}{2} \tan ^{-1} x-\int \frac{1}{1+x^2} \times \frac{x^2}{2} d x\right]+c \\
\mathrm{I}=2\left[\frac{x^2}{2} \tan ^{-1} x-\frac{1}{2} \int \frac{x^2+1-1}{x^2+1} d x\right]+c \\
\mathrm{I}=2\left[\frac{x^2}{2} \tan ^{-1} x-\frac{1}{2} \int \frac{x^2+1}{x^2+1} d x+\frac{1}{2} \int \frac{1}{1+x^2} d x\right]+c
\end{array}
$$

$$
\begin{aligned}
\mathrm{I} &=2\left[\frac{x^2}{2} \tan ^{-1} x-\frac{1}{2} \int 1 . d x+\frac{1}{2} \tan ^{-1} x\right]+c \\
\mathrm{I} &=2\left[\frac{x^2}{2} \tan ^{-1} x-\frac{x}{2}+\frac{1}{2} \tan ^{-1} x\right]+c \\
\mathrm{I} &=x^2 \tan ^{-1} x+\tan ^{-1} x-x+c \\
& \text { or } I=-x+\left(x^2+1\right) \tan ^{-1} x+c
\end{aligned}
$$