Answer
$\frac{1}{4}x\left( {{x^2} + 2} \right)\sqrt {{x^2} + 4} - 2{\sinh ^{ - 1}}\left( {\frac{x}{2}} \right) + C$
Work Step by Step
$$\eqalign{
& \int {{x^2}\sqrt {{x^2} + 4} } dx \cr
& {\text{Integrate by using a CAS }}\left( {{\text{See image below}}} \right){\text{ we obtain:}} \cr
& \int {{x^2}\sqrt {{x^2} + 4} } dx = \frac{1}{4}x\left( {{x^2} + 2} \right)\sqrt {{x^2} + 4} - 2{\sinh ^{ - 1}}\left( {\frac{x}{2}} \right) + C \cr
& \cr
& {\text{Integrating by tables using the entry 22}} \cr
& \int {{u^2}\sqrt {{u^2} + {a^2}} du = \frac{u}{8}\left( {{a^2} + 2{u^2}} \right)\sqrt {{a^2} + {u^2}} - \frac{{{a^4}}}{8}\ln \left( {u + \sqrt {{u^2} + {a^2}} } \right)} \cr
& + C \cr
& \cr
& {\text{Therefore}}{\text{, for }}\int {{x^2}\sqrt {{x^2} + 4} } dx{\text{ we obtain:}} \cr
& = \frac{x}{8}\left( {4 + 2{x^2}} \right)\sqrt {{x^2} + 4} - \frac{{{{\left( 2 \right)}^4}}}{8}\ln \left( {x + \sqrt {{x^2} + 4} } \right) + C \cr
& {\text{Simplifying}} \cr
& = \frac{{2x}}{8}\left( {{x^2} + 2} \right)\sqrt {{x^2} + 4} - 2\ln \left( {x + \sqrt {{x^2} + 4} } \right) + C \cr
& = \frac{1}{4}x\left( {{x^2} + 2} \right)\sqrt {{x^2} + 4} - 2\ln \left( {x + \sqrt {{x^2} + 4} } \right) + C \cr
& {\text{Where }}{\sinh ^{ - 1}}\left( u \right) = \ln \left( {u + \sqrt {{u^2} + 1} } \right),{\text{ then}} \cr
& = \frac{1}{4}x\left( {{x^2} + 2} \right)\sqrt {{x^2} + 4} - 2{\sinh ^{ - 1}}\left( {\frac{x}{2}} \right) + C \cr
& {\text{We obtain the same result.}} \cr} $$
