Answer
$$\frac{x}{2}\sqrt {{x^2} + 4} - 2\ln \left( {x + \sqrt {{x^2} + 4} } \right) + C\,\,\,$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^2}}}{{\sqrt {{x^2} + 4} }}} dx \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral}} \cr
& {\text{Rewrite the integrand}} \cr
& \int {\frac{{{x^2}}}{{\sqrt {{x^2} + 4} }}} dx = \int {\frac{{{x^2}}}{{\sqrt {{x^2} + {{\left( 2 \right)}^2}} }}} dx \cr
& {\text{The integrand has a expression in the form }}\sqrt {{u^2} + {a^2}} {} \cr
& {\text{Use formula 95}} \cr
& \left( {95} \right):\,\,\,\,\int {\frac{{{u^2}}}{{\sqrt {{u^2} + {a^2}} }}du} = \frac{u}{2}\sqrt {{u^2} + {a^2}} - \frac{{{a^2}}}{2}\ln \left( {u + \sqrt {{u^2} + {a^2}} } \right) + C\,\,\, \cr
& {\text{let }}u = x,\,\,\,a = 2 \cr
& \int {\frac{{{x^2}}}{{\sqrt {{x^2} + {{\left( 2 \right)}^2}} }}} dx = \frac{x}{2}\sqrt {{x^2} + {{\left( 2 \right)}^2}} - \frac{{{{\left( 2 \right)}^2}}}{2}\ln \left( {x + \sqrt {{x^2} + {{\left( 2 \right)}^2}} } \right) + C\,\,\, \cr
& {\text{simplifying}} \cr
& \int {\frac{{{x^2}}}{{\sqrt {{x^2} + 4} }}} dx = \frac{x}{2}\sqrt {{x^2} + 4} - 2\ln \left( {x + \sqrt {{x^2} + 4} } \right) + C\,\,\, \cr} $$
