Answer
$$\frac{{{{\left( {1 + {x^2}} \right)}^{3/2}}}}{3} - \sqrt {1 + {x^2}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^3}}}{{\sqrt {1 + {x^2}} }}} dx \cr
& {\text{Taking the hint on the book making a substitution of the form }}u = {\left( {x + a} \right)^{1/n}} \cr
& {\text{Take }}\,\,\,u = {\left( {1 + {x^2}} \right)^{1/2}} \to \,\,\,\,\,\,\,\,{u^2} = 1 + {x^2},\,\,\,\,\,\,\,2udu = 2xdx\,\,\,\, \cr
& \cr
& {\text{write the integrand in terms of }}u \cr
& \int {\frac{{{x^3}}}{{\sqrt {1 + {x^2}} }}} dx = \int {\frac{{{x^3}}}{u}} \left( {\frac{{2udu}}{{2x}}} \right) \cr
& = \int {{x^2}} du \cr
& = \int {\left( {{u^2} - 1} \right)} du \cr
& \cr
& {\text{integrating}} \cr
& = \frac{{{u^3}}}{3} - u + C \cr
& \cr
& {\text{write the integrand in terms of }}x;{\text{ replace }}u = {\left( {1 + {x^2}} \right)^{1/2}} \cr
& = \frac{{{{\left( {{{\left( {1 + {x^2}} \right)}^{1/2}}} \right)}^3}}}{3} - {\left( {1 + {x^2}} \right)^{1/2}} + C \cr
& = \frac{{{{\left( {1 + {x^2}} \right)}^{3/2}}}}{3} - \sqrt {1 + {x^2}} + C \cr} $$
