Answer
$$x\arcsin \left( {4x} \right) + \frac{1}{4}\sqrt {1 - 16{x^2}} + C$$
Work Step by Step
$$\eqalign{
& \int {\arcsin 4x} dx \cr
& = \frac{1}{4}\int {\arcsin 4x} \left( 4 \right)dx \cr
& u = 4x,{\text{ }}du = 4dx \cr
& = \frac{1}{4}\int {\arcsin u} du \cr
& {\text{From the table of integrals in the back of the book}} \cr
& \int {\arcsin u} du = u\arcsin u + \sqrt {1 - {u^2}} + C \cr
& \frac{1}{4}\int {\arcsin u} du = \frac{1}{4}u\arcsin u + \frac{1}{4}\sqrt {1 - {u^2}} + C \cr
& {\text{Write in terms of }}x,{\text{ }}u = 4x \cr
& = \frac{1}{4}\left( {4x} \right)\arcsin \left( {4x} \right) + \frac{1}{4}\sqrt {1 - {{\left( {4x} \right)}^2}} + C \cr
& = x\arcsin \left( {4x} \right) + \frac{1}{4}\sqrt {1 - 16{x^2}} + C \cr} $$
