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Let $I=\int \frac{x \cos ^{-1} x}{\sqrt{1-x^{2}}} d x$

$I=\frac{-1}{2} \int \frac{-2 x}{\sqrt{1-x^{2}}} \cdot \cos ^{-1} x d x$

Taking $\cos ^{-1} x$ as first function and $\left(\frac{-2 x}{\sqrt{1-x^{2}}}\right)$ as second function and integrating by parts, we obtain

$\begin{aligned} I &=\frac{-1}{2}\left[\cos ^{-1} x \int \frac{-2 x}{\sqrt{1-x^{2}}} d x-\int\left\{\left(\frac{d}{d x} \cos ^{-1} x\right) \int \frac{-2 x}{\sqrt{1-x^{2}}} d x\right\} d x\right] \\ &=\frac{-1}{2}\left[\cos ^{-1} x \cdot 2 \sqrt{1-x^{2}}-\int \frac{-1}{\sqrt{1-x^{2}}} \cdot 2 \sqrt{1-x^{2}} d x\right] \\ &=\frac{-1}{2}\left[2 \sqrt{1-x^{2}} \cos ^{-1} x+\int 2 d x\right] \\ &=\frac{-1}{2}\left[2 \sqrt{1-x^{2}} \cos ^{-1} x+2 x\right]+\mathrm{C} \\ &=-\left[\sqrt{1-x^{2}} \cos ^{-1} x+x\right]+\mathrm{C} \end{aligned}$