Answer
$$ - \left( {{x^2} + 2} \right)\sqrt {1 - {x^2}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{3{x^3}}}{{\sqrt {1 - {x^2}} }}} dx \cr
& {\text{substitute }}x = \sin \theta ,{\text{ }}dx = \cos \theta d\theta \cr
& = \int {\frac{{3{{\left( {\sin \theta } \right)}^3}}}{{\sqrt {1 - {{\sin }^2}\theta } }}} \left( {\cos \theta } \right)d\theta \cr
& {\text{simplify}} \cr
& = \int {\frac{{3{{\sin }^3}\theta \cos \theta }}{{\sqrt {{{\cos }^2}\theta } }}d\theta } \cr
& = \int {\frac{{3{{\sin }^3}\theta \cos \theta }}{{\cos \theta }}} d\theta \cr
& = 3\int {{{\sin }^3}\theta } d\theta \cr
& = 3\int {\left( {{{\sin }^2}\theta } \right)\sin \theta } d\theta \cr
& {\text{use si}}{{\text{n}}^2}x + {\cos ^2}x = 1 \cr
& = 3\int {\left( {1 - {{\cos }^2}\theta } \right)\sin \theta } d\theta \cr
& = 3\int {\left( {\sin \theta - {{\cos }^2}\theta \sin \theta } \right)} d\theta \cr
& {\text{integrating}} \cr
& = 3\left( { - \cos \theta + \frac{{{{\cos }^3}\theta }}{3}} \right) + C \cr
& = - 3\cos \theta + {\cos ^3}\theta + C \cr
& {\text{writing the answer in terms of }}x,{\text{ we have that }} \cr
& x = \sin \theta ,{\text{ then }}\cos \theta = \sqrt {1 - {x^2}} \cr
& = - 3\sqrt {1 - {x^2}} + {\left( {\sqrt {1 - {x^2}} } \right)^3} + C \cr
& {\text{factoring}} \cr
& = \sqrt {1 - {x^2}} \left( { - 3 + {{\left( {\sqrt {1 - {x^2}} } \right)}^2}} \right) + C \cr
& = \sqrt {1 - {x^2}} \left( { - 3 + 1 - {x^2}} \right) + C \cr
& = \sqrt {1 - {x^2}} \left( { - 2 - {x^2}} \right) + C \cr
& = - \left( {{x^2} + 2} \right)\sqrt {1 - {x^2}} + C \cr} $$
