Answer
$$\frac{x^{2}}{2} \cos ^{-1} x+\frac{1}{4} \sin ^{-1} x-\frac{1}{4} x \sqrt{1-x^{2}}+C$$
Work Step by Step
By using the formula
$$ \int x^{n} \cos ^{-1} a x d x=\frac{x^{n+1}}{n+1} \cos ^{-1} a x+\frac{a}{n+1} \int \frac{x^{n+1} d x}{\sqrt{1-a^{2} x^{2}}}, \quad n \neq-1$$
Here $n=1$, we get
\begin{align*}
\int x \cos ^{-1} x d x&=\int x^{1} \cos ^{-1} x d x\\
&=\frac{x^{1+1}}{1+1} \cos ^{-1} x+\frac{1}{1+1} \int \frac{x^{1+1} d x}{\sqrt{1-x^{2}}}\\
&=\frac{x^{2}}{2} \cos ^{-1} x+\frac{1}{2} \int \frac{x^{2} d x}{\sqrt{1-x^{2}}}
\end{align*}
Then use
$$\int \frac{x^{2}}{\sqrt{a^{2}-x^{2}}} d x=\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}-\frac{1}{2} x \sqrt{a^{2}-x^{2}}+C $$
with $a=1$, we get
\begin{align*}
\int x \cos ^{-1} x d x &=\frac{x^{2}}{2} \cos ^{-1} x+\frac{1}{2} \int \frac{x^{2} d x}{\sqrt{1-x^{2}}}\\
&=\frac{x^{2}}{2} \cos ^{-1} x+\frac{1}{2}\left(\frac{1}{2} \sin ^{-1} x\right)-\frac{1}{2}\left(\frac{1}{2} x \sqrt{1-x^{2}}\right)+C\\
&=\frac{x^{2}}{2} \cos ^{-1} x+\frac{1}{4} \sin ^{-1} x-\frac{1}{4} x \sqrt{1-x^{2}}+C
\end{align*}
