Answer
\[ = \frac{1}{{10}}{\tan ^{ - 1}}\,\left( {\frac{x}{5}} \right) - \frac{x}{{2\,\left( {25 + {x^2}} \right)}} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{{x^2}}}{{\,{{\left( {25 + {x^2}} \right)}^2}}}} dx \hfill \\
\hfill \\
rewrite \hfill \\
\hfill \\
= \int_{}^{} {\frac{{{x^2}}}{{\,\left( {\sqrt {25 + {x^2}} } \right)}}dx} \hfill \\
\hfill \\
the\,integral\,contains\,\,the\,\,form\,\,{a^2} + {x^2}\,\,so. \hfill \\
\hfill \\
x = 5\tan \theta \,\,\,\,\,then\,\,\,\,dx = 5{\sec ^2}\theta d\theta \hfill \\
and\,\,\sqrt {25 + {x^2}} = 5\sec \theta \hfill \\
\hfill \\
Substituting{\text{ }}these{\text{ }}factors{\text{ }} \hfill \\
\hfill \\
= \int_{}^{} {\frac{{\,\left( {25{{\tan }^2}\theta } \right)\,\left( {5{{\sec }^2}\theta } \right)d\theta }}{{\,{{\left( {5\sec \theta } \right)}^4}}}} \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
= \frac{1}{5}\int_{}^{} {\frac{{{{\tan }^2}\theta }}{{{{\sec }^2}\theta }}d\theta } = \frac{1}{5}\int_{}^{} {si{n^2}\theta d\theta } \hfill \\
\hfill \\
use\,the\,trigonometric\,\,identity{\text{ }}{\sin ^2}\theta = \frac{{1 - \cos \theta }}{2} \hfill \\
\hfill \\
= \frac{1}{5}\int_{}^{} {\,\left( {\frac{{1 - \cos 2\theta }}{2}} \right)} d\theta \hfill \\
\hfill \\
integrate \hfill \\
\hfill \\
= \frac{1}{5}\,\left( {\frac{\theta }{2} - \frac{{\sin \theta \cos \theta }}{2}} \right) + C \hfill \\
\hfill \\
= \frac{1}{{10}}{\tan ^{ - 1}}\,\left( {\frac{x}{5}} \right) - \frac{1}{{10}}\,\left( {\frac{x}{{\sqrt {25 + {x^2}} }}} \right)\,\left( {\frac{5}{{\sqrt {25 + {x^2}} }}} \right) + C \hfill \\
\hfill \\
substitute\,\,for\,\,\sin \theta {\text{ and cos}}\theta \hfill \\
\hfill \\
= \frac{1}{{10}}{\tan ^{ - 1}}\,\left( {\frac{x}{5}} \right) - \frac{x}{{2\,\left( {25 + {x^2}} \right)}} + C \hfill \\
\hfill \\
\end{gathered} \]
