Answer
\[ = - \frac{1}{x}{\tan ^{ - 1}}x + \frac{1}{2}\ln \left| {\frac{{{x^2}}}{{1 + {x^2}}}} \right| + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{{{\tan }^{ - 1}}\,x}}{{{x^2}}}\,dx} \hfill \\
\hfill \\
Integrate\,\,by\,\,tables,\,\,use\,\,the\,\,formulas \hfill \\
\hfill \\
\int_{}^{} {{x^n}{{\tan }^{ - 1}}xdx} = \frac{1}{{n + 1}}\,\,\left[ {{x^{n + 1}}{{\tan }^{ - 1}}x - \int_{}^{} {\frac{{{x^{n + 1}}}}{{1 + {x^2}}}\,dx} } \right] + C \hfill \\
\hfill \\
\,\,\int_{}^{} {\frac{1}{{x\,\left( {{x^2} + {a^2}} \right)}}dx} = \frac{1}{{2{a^2}}}\ln \left| {\frac{{{x^2}}}{{{a^2} + {x^2}}}} \right| + C \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
\int_{}^{} {{x^{ - 2}}{{\tan }^{ - 1}}xdx} = \,\,\left[ {\frac{1}{x}{{\tan }^{ - 1}}x - \int_{}^{} {\frac{1}{{x\,\left( {1 + {x^2}} \right)}}dx} } \right] \hfill \\
\hfill \\
\int_{}^{} {{x^{ - 2}}{{\tan }^{ - 1}}xdx} = - \frac{1}{x}{\tan ^{ - 1}}x + \frac{1}{2}\ln \left| {\frac{{{x^2}}}{{1 + {x^2}}}} \right| + C \hfill \\
\hfill \\
\,\,using\,\,a\,\,\,CAS\,\,we\,\,get\,\,the\,\,same\,\,result \hfill \\
\hfill \\
\int_{}^{} {{x^{ - 2}}{{\tan }^{ - 1}}xdx} = - \frac{1}{x}{\tan ^{ - 1}}x + \frac{1}{2}\ln \left| {\frac{{{x^2}}}{{1 + {x^2}}}} \right| + C \hfill \\
\hfill \\
\end{gathered} \]
