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

(N/A) Let the first function be $u = \sin ^{-1} x$ and the second function be $dv = \frac{x}{\sqrt{1-x^{2}}} dx$.
First,we find the integral of the second function:
$\int \frac{x}{\sqrt{1-x^{2}}} dx$
Let $t = 1-x^{2}$,then $dt = -2x dx$,which implies $x dx = -\frac{1}{2} dt$.
$\int \frac{x}{\sqrt{1-x^{2}}} dx = -\frac{1}{2} \int t^{-1/2} dt = -\frac{1}{2} (2t^{1/2}) = -\sqrt{1-x^{2}}$.
Now,apply the integration by parts formula: $\int u dv = uv - \int v du$.
Here,$du = \frac{1}{\sqrt{1-x^{2}}} dx$.
$\int \frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}} dx = (\sin ^{-1} x)(-\sqrt{1-x^{2}}) - \int (-\sqrt{1-x^{2}}) \frac{1}{\sqrt{1-x^{2}}} dx$
$= -\sqrt{1-x^{2}} \sin ^{-1} x + \int 1 dx$
$= -\sqrt{1-x^{2}} \sin ^{-1} x + x + C$
$= x - \sqrt{1-x^{2}} \sin ^{-1} x + C$.