Evaluate the following integrals:

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

We are splitting this in to two functions

First we find the integral of:

$\int \frac{\mathrm{x}}{\sqrt{1-\mathrm{x}^{2}}} \mathrm{dx}$

Put $1-x^{2}=t$

$-2 x d x=d t$

$\int \frac{\mathrm{x}}{\sqrt{1-\mathrm{x}^{2}}} \mathrm{dx}=-\frac{1}{2} \int \frac{\mathrm{dt}}{\sqrt{\mathrm{t}}}=-\sqrt{\mathrm{t}}=-\sqrt{1-\mathrm{x}^{2}}$

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

Using integration by parts,

$=\left(\sin ^{-1} x\right) \times-\sqrt{1-x^{2}}-\int \frac{1}{\sqrt{1-x^{2}}}\left(-\sqrt{1-x^{2}}\right) d x$

$=\left(\sin ^{-1} x\right) \times-\sqrt{1-x^{2}}-\int d x$

$=\left(\sin ^{-1} x\right) x-\sqrt{1-x^{2}}+x+c$

$=x-\sqrt{1-x^{2}}\left(\sin ^{-1} x\right)+c$