Answer
$$ x - 2{\tan ^{ - 1}}\left( {\frac{x}{2}} \right) + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^2}}}{{4 + {x^2}}}} dx \cr
& {\text{Using Trigonometric Substitutions}} \cr
& {\text{We set}}{\text{, }}x = 2\tan \theta,{\text{ }}dx = 2{\sec ^2}\theta d\theta \cr
& {\text{With these substitutions}}{\text{, we have}} \cr
& \int {\frac{{{x^2}}}{{4 + {x^2}}}} dx = \int {\frac{{{{\left( {2\tan \theta } \right)}^2}}}{{4 + {{\left( {2\tan \theta } \right)}^2}}}} \left( {2{{\sec }^2}\theta } \right)d\theta \cr
& = \int {\frac{{4{{\tan }^2}\theta }}{{4 + 4{{\tan }^2}\theta }}} \left( {2{{\sec }^2}\theta } \right)d\theta \cr
& = \int {\frac{{{{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \left( {2{{\sec }^2}\theta } \right)d\theta \cr
& {\text{use the fundamental identity }}{\sec ^2}\theta - 1 = {\tan ^2}\theta \cr
& = \int {\frac{{{{\tan }^2}\theta }}{{{{\sec }^2}\theta }}} \left( {{{\sec }^2}\theta } \right)d\theta \cr
& = 2\int {{{\tan }^2}\theta } d\theta \cr
& = 2\int {\left( {{{\sec }^2}\theta - 1} \right)} d\theta \cr
& {\text{integrating}} \cr
& = 2\tan \theta - 2\theta + C \cr
& {\text{write in terms of }}x,{\text{ }}\tan \theta = \frac{x}{2}{\text{ and }}\theta = {\tan ^{ - 1}}\theta \cr
& = 2\left( {\frac{x}{2}} \right) - 2{\tan ^{ - 1}}\left( {\frac{x}{2}} \right) + C \cr
& = x - 2{\tan ^{ - 1}}\left( {\frac{x}{2}} \right) + C \cr} $$
