Integrate the function: $x \sin^{-1} x$

Let $I = \int x \sin^{-1} x \, dx$.
Using integration by parts,where $\sin^{-1} x$ is the first function and $x$ is the second function:
$I = \sin^{-1} x \int x \, dx - \int \left( \frac{d}{dx} \sin^{-1} x \int x \, dx \right) dx$
$I = \sin^{-1} x \left( \frac{x^2}{2} \right) - \int \frac{1}{\sqrt{1-x^2}} \cdot \frac{x^2}{2} \, dx$
$I = \frac{x^2 \sin^{-1} x}{2} - \frac{1}{2} \int \frac{x^2}{\sqrt{1-x^2}} \, dx$
To solve the integral,rewrite the numerator as $x^2 = -(1-x^2) + 1$:
$I = \frac{x^2 \sin^{-1} x}{2} - \frac{1}{2} \int \frac{-(1-x^2) + 1}{\sqrt{1-x^2}} \, dx$
$I = \frac{x^2 \sin^{-1} x}{2} + \frac{1}{2} \int \sqrt{1-x^2} \, dx - \frac{1}{2} \int \frac{1}{\sqrt{1-x^2}} \, dx$
Using standard integrals $\int \sqrt{a^2-x^2} \, dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a})$ and $\int \frac{1}{\sqrt{1-x^2}} \, dx = \sin^{-1} x$:
$I = \frac{x^2 \sin^{-1} x}{2} + \frac{1}{2} \left( \frac{x}{2} \sqrt{1-x^2} + \frac{1}{2} \sin^{-1} x \right) - \frac{1}{2} \sin^{-1} x + C$
$I = \frac{x^2 \sin^{-1} x}{2} + \frac{x}{4} \sqrt{1-x^2} + \frac{1}{4} \sin^{-1} x - \frac{1}{2} \sin^{-1} x + C$
$I = \frac{1}{4} (2x^2 - 1) \sin^{-1} x + \frac{x}{4} \sqrt{1-x^2} + C$,where $C$ is an arbitrary constant.