Answer
$$\frac{-1}{x}\sin^{-1}x-\ln | \frac{1}{x}+\frac{\sqrt{1-x^2}}{x}|+c $$
Work Step by Step
Given
$$ \int \frac{\sin^{-1}x}{x^2}dx $$
Use integration by parts , let
\begin{aligned}
u&= \sin^{-1}x\ \ \ \ \ \ \ &dv&= x^{-2}dx\\
du&= \frac{dx}{\sqrt{1-x^2}}\ \ \ \ \ \ \ & v&= \frac{-1}{x}
\end{aligned}
Then
\begin{aligned}
\int \frac{\sin^{-1}x}{x^2}dx&= \frac{-1}{x}\sin^{-1}x+\int \frac{dx}{x\sqrt{1-x^2}}
\end{aligned}
To evaluate $\int \dfrac{dx}{x\sqrt{1-x^2}}$, let
$$x=\sin \theta \ \ \ \ \ dx=\cos \theta d\theta $$
Then
\begin{aligned}
\int \dfrac{dx}{x\sqrt{1-x^2}}&=\int \frac{\cos \theta d\theta }{\sin \theta \sqrt{1-\sin^2\theta }}\\
&=\int \frac{1}{\sin \theta } d\theta \\
&=\int \csc \theta d\theta \\
&=-\ln |\csc \theta +\cot \theta |+c\\
&= -\ln | \frac{1}{x}+\frac{\sqrt{1-x^2}}{x}|+c
\end{aligned}
Hence
\begin{aligned}
\int \frac{\sin^{-1}x}{x^2}dx&= \frac{-1}{x}\sin^{-1}x-\ln | \frac{1}{x}+\frac{\sqrt{1-x^2}}{x}|+c
\end{aligned}
