Integrate the function: $\frac{x \cos^{-1} x}{\sqrt{1-x^{2}}}$

Let $I = \int \frac{x \cos^{-1} x}{\sqrt{1-x^{2}}} dx$.
We can rewrite the integral as $I = -\frac{1}{2} \int \frac{-2x}{\sqrt{1-x^{2}}} \cdot \cos^{-1} x dx$.
Using integration by parts,let $u = \cos^{-1} x$ and $dv = \frac{-2x}{\sqrt{1-x^{2}}} dx$.
Then $du = -\frac{1}{\sqrt{1-x^{2}}} dx$ and $v = 2\sqrt{1-x^{2}}$.
Using the formula $\int u dv = uv - \int v du$:
$I = -\frac{1}{2} \left[ \cos^{-1} x \cdot 2\sqrt{1-x^{2}} - \int 2\sqrt{1-x^{2}} \cdot \left( -\frac{1}{\sqrt{1-x^{2}}} \right) dx \right]$.
$I = -\frac{1}{2} \left[ 2\sqrt{1-x^{2}} \cos^{-1} x + \int 2 dx \right]$.
$I = -\frac{1}{2} \left[ 2\sqrt{1-x^{2}} \cos^{-1} x + 2x \right] + C$.
$I = -\sqrt{1-x^{2}} \cos^{-1} x - x + C$,where $C$ is the constant of integration.