Answer
$$\frac{1}{2}\left( {{{\tan }^{ - 1}}x - \frac{x}{{1 + {x^2}}}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^2}}}{{{{\left( {1 + {x^2}} \right)}^2}}}} dx \cr
& {\text{Using substitution }}x = \tan \theta \cr
& x = \tan \theta ,{\text{ }}dx = {\sec ^2}\theta d\theta \cr
& {\text{Substitute}} \cr
& \int {\frac{{{x^2}}}{{{{\left( {1 + {x^2}} \right)}^2}}}} dx = \int {\frac{{{{\tan }^2}\theta }}{{{{\left( {1 + {{\tan }^2}\theta } \right)}^2}}}{{\sec }^2}\theta } d\theta \cr
& {\text{Use the pythagorean identity }}{\tan ^2}\theta + 1 = {\sec ^2}\theta \cr
& = \int {\frac{{{{\tan }^2}\theta }}{{{{\left( {{{\sec }^2}\theta } \right)}^2}}}{{\sec }^2}\theta } d\theta \cr
& = \int {\frac{{{{\tan }^2}\theta }}{{{{\sec }^2}\theta }}} d\theta = \int {\left( {\frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}} \right)\left( {\frac{{{{\cos }^2}}}{1}} \right)} d\theta \cr
& = \int {{{\sin }^2}\theta } d\theta \cr
& {\text{Use the power reducing formula }}{\sin ^2}u = \frac{{1 - \cos 2u}}{2} \cr
& = \int {\left( {\frac{1}{2} - \frac{1}{2}\cos 2\theta } \right)} d\theta \cr
& {\text{Integrating}} \cr
& = \frac{1}{2}\theta - \frac{1}{2}\left( {\frac{1}{2}\sin 2\theta } \right) + C \cr
& {\text{Use the double - angle formula sin2}}\theta {\text{ = 2sin}}\theta {\text{cos}}\theta \cr
& = \frac{1}{2}\theta - \frac{1}{2}\left( {\frac{1}{2}\left( {{\text{2sin}}\theta {\text{cos}}\theta } \right)} \right) + C \cr
& = \frac{1}{2}\theta - \frac{1}{2}{\text{sin}}\theta {\text{cos}}\theta + C \cr
& {\text{Where,}} \cr
& \tan \theta = x,{\text{ }}\tan \theta = \frac{x}{1},{\text{ sin}}\theta = \frac{x}{{\sqrt {1 + {x^2}} }},\cos \theta = \frac{1}{{\sqrt {1 + {x^2}} }} \cr
& = \frac{1}{2}{\tan ^{ - 1}}x - \frac{1}{2}\left( {\frac{x}{{\sqrt {1 + {x^2}} }}} \right)\left( {\frac{1}{{\sqrt {1 + {x^2}} }}} \right) + C \cr
& = \frac{1}{2}{\tan ^{ - 1}}x - \frac{x}{{2\left( {1 + {x^2}} \right)}} + C \cr
& {\text{Factoring}} \cr
& = \frac{1}{2}\left( {{{\tan }^{ - 1}}x - \frac{x}{{1 + {x^2}}}} \right) + C \cr} $$
