Answer
$$\frac{1}{4}{\arcsin ^2}2x + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\arcsin 2x}}{{\sqrt {1 - 4{x^2}} }}} dx \cr
& = \int {\arcsin 2x\left( {\frac{1}{{\sqrt {1 - 4{x^2}} }}} \right)} dx \cr
& {\text{Integrate by substitution}} \cr
& u = \arcsin 2x,{\text{ }}du = \frac{1}{{\sqrt {1 - {{\left( {2x} \right)}^2}} }}\left( 2 \right)dx \cr
& du = \frac{1}{{\sqrt {1 - 4{x^2}} }}2dx,{\text{ }}\frac{1}{2}du = \frac{1}{{\sqrt {1 - 4{x^2}} }}dx \cr
& {\text{Substituting}} \cr
& \int {\arcsin 2x\left( {\frac{1}{{\sqrt {1 - 4{x^2}} }}} \right)} dx = \int {u\left( {\frac{1}{2}} \right)} du \cr
& = \frac{1}{2}\int u du \cr
& = \frac{1}{4}{u^2} + C \cr
& {\text{Write in terms of }}x \cr
& = \frac{1}{4}{\left( {\arcsin 2x} \right)^2} + C \cr
& = \frac{1}{4}{\arcsin ^2}2x + C \cr} $$
