Find ∫ (sin^-1 x)/((1-x^2)^3/2) dx

The correct answer is Option (4) → $\frac{x}{\sqrt{1-x^2}} \sin^{-1} x + \log |\sqrt{1-x^2}| + C$

$I = \int \frac{\sin^{-1} x}{(1-x^2)^{3/2}} dx$

Let $\sin^{-1} x = t$

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

$I = \int \frac{t}{(1-\sin^2 t)} dt = \int t \sec^2 t dt$

$I = t \int \sec^2 t dt - \int \left[ \frac{d}{dt} t \int \sec^2 t dt \right]$

$= t \tan t - \int \tan t dt$

$= t \tan t + \log |\cos t| + C$

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