$∫\frac{xsin^{-1}x}{\sqrt{1-x^2}}dx=$

$-\sqrt{1-x^2}sin^{-1} \, x+ x +C$ where C is arbitrary constant

The correct answer is Option (4) → $-\sqrt{1-x^2}sin^{-1} \, x+ x +C$ where C is arbitrary constant

$∫\frac{x\sin^{-1}x}{\sqrt{1-x^2}}dx$

let $y=\sin^{-1}x⇒\sin y =x$

$dy=\frac{dx}{\sqrt{1-x^2}}$

$\int y\sin y dy$

Using $\int uvdx=u\int vdx=\int u'\int v dxdx$

$=y\int \sin ydy-\int\frac{d(y)}{dy}\int\sin ydydy$

$=-y\cos y+\int\cos y dy$

$=-y\cos y+\sin y+C$

$≡-\sqrt{1-x^2}\sin^{-1}+x+C$