Answer
$$ - \frac{1}{x}{\tan ^{ - 1}}x + \ln \left| x \right| - \frac{1}{2}\ln \left( {{x^2} + 1} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{{\tan }^{ - 1}}x}}{{{x^2}}}} dx \cr
& {\text{Integrate by parts }} \cr
& {\text{Let }}u = {\tan ^{ - 1}}x,{\text{ }}du = \frac{1}{{{x^2} + 1}}dx \cr
& dv = \frac{1}{{{x^2}}},{\text{ }}v = - \frac{1}{x} \cr
& {\text{Use integration by parts formula}} \cr
& \int {udv} = uv - \int {vdu} \cr
& \int {\frac{{{{\tan }^{ - 1}}x}}{{{x^2}}}} dx = - \frac{1}{x}{\tan ^{ - 1}}x - \int {\left( { - \frac{1}{x}} \right)} \left( {\frac{1}{{{x^2} + 1}}} \right)dx \cr
& \int {\frac{{{{\tan }^{ - 1}}x}}{{{x^2}}}} dx = - \frac{1}{x}{\tan ^{ - 1}}x + \int {\frac{1}{{x\left( {{x^2} + 1} \right)}}} dx{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Decomposing }}\frac{1}{{x\left( {{x^2} + 1} \right)}}{\text{ into partial fractions}} \cr
& \frac{1}{{x\left( {{x^2} + 1} \right)}} = \frac{1}{x} - \frac{x}{{{x^2} + 1}},{\text{ substuting into }}\left( {\bf{1}} \right) \cr
& \int {\frac{{{{\tan }^{ - 1}}x}}{{{x^2}}}} dx = - \frac{1}{x}{\tan ^{ - 1}}x + \int {\left( {\frac{1}{x} - \frac{x}{{{x^2} + 1}}} \right)} dx \cr
& \int {\frac{{{{\tan }^{ - 1}}x}}{{{x^2}}}} dx = - \frac{1}{x}{\tan ^{ - 1}}x + \int {\frac{1}{x}} dx - \int {\frac{x}{{{x^2} + 1}}} dx \cr
& {\text{Integrating}} \cr
& \int {\frac{{{{\tan }^{ - 1}}x}}{{{x^2}}}} dx = - \frac{1}{x}{\tan ^{ - 1}}x + \ln \left| x \right| - \frac{1}{2}\ln \left( {{x^2} + 1} \right) + C \cr} $$