Answer
$$-\sqrt{1-x^2} + \frac{1}{2}(\arcsin x)^2+c $$
Work Step by Step
Given
$$\int \frac{x+\arcsin x}{\sqrt{1-x^2}} d x$$
Since
\begin{aligned}
\int \frac{x+\arcsin x}{\sqrt{1-x^2}} d x&=\int \frac{x }{\sqrt{1-x^2}} d x+\int \frac{ \arcsin x}{\sqrt{1-x^2}} d x\\
&= \frac{-1}{2}\int -2x (1-x^2)^{-1/2} d x+\int \frac{ \arcsin x}{\sqrt{1-x^2}} d x\\
&= -\sqrt{1-x^2} + \frac{1}{2}(\arcsin x)^2+c
\end{aligned}
