$\int \frac{\sqrt{1-x^2} \sin ^{-1} x+x}{\sqrt{1-x^2}} d x=$
$\int \frac{\sqrt{1-x^2} \sin ^{-1} x+x}{\sqrt{1-x^2}} d x=$
$x \sin ^{-1} x+\sqrt{1-x^2}+c$
$\sin ^{-1} x+\sqrt{1-x^2}+c$
$x \sin ^{-1} x+c$
$\frac{x \sin ^{-1} x}{\sqrt{1-x^2}}+c$
Solution
$
\begin{aligned}
& \text { Given, } \int \frac{\sqrt{1-x^2} \cdot \sin ^{-1} x+x}{\sqrt{1-x^2}} d x \\
& \int\left(\frac{\sqrt{1-x^2} \cdot \sin ^{-1} x}{\sqrt{1-x^2}}+\frac{x}{\sqrt{1-x^2}}\right) d x \\
& \int\left(\sin ^{-1} x+\frac{x}{\sqrt{1-x^2}}\right) d x \\
& \int \sin ^{-1} x d x+\int \frac{x}{\sqrt{1-x^2}} d x \\
& \sin ^{-1} x \cdot \int 1 d x-\int\left(\frac{d}{d x} \sin ^{-1} x \cdot \int 1 d x\right) d x+\int \frac{x}{\sqrt{1-x^2}} d x \\
& \quad\left[\because \text { By using by parts with } f(x)=\sin ^{-1} x ; g(x)=1\right] \\
& \sin ^{-1} x \cdot x-\int \frac{1}{\sqrt{1-x^2}} \cdot x d x+\int \frac{x}{\sqrt{1-x^2}} d x+c \\
& x \cdot \sin ^{-1} x+c
\end{aligned}
$
$\therefore$ Hence option (c) is correct