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

$x - \sqrt{1 - x^2} \sin^{-1} x + C$

The correct answer is Option (1) → $x - \sqrt{1 - x^2} \sin^{-1} x + C$

Let first function be $\sin^{-1} x$ and second function be $\frac{x}{\sqrt{1-x^2}}$.

First we find the integral of the second function, i.e., $\int \frac{x \, dx}{\sqrt{1-x^2}}$.

Put $t = 1 - x^2$. Then $dt = -2x \, dx$

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

Hence, $\int \frac{x \sin^{-1} x}{\sqrt{1-x^2}} \, dx = (\sin^{-1} x) (-\sqrt{1-x^2}) - \int \frac{1}{\sqrt{1-x^2}} (-\sqrt{1-x^2}) \, dx$

$= -\sqrt{1-x^2} \sin^{-1} x + x + C = x - \sqrt{1-x^2} \sin^{-1} x + C$