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
$$\eqalign{
& \int {\frac{{{{\tan }^{ - 1}}x}}{{{x^2}}}} dx \cr
& = \int {\frac{1}{{{x^2}}}{{\tan }^{ - 1}}x} dx \cr
& {\text{Using the method of integration by parts}} \cr
& \,\,\,\,\,\,\,\,\,{\text{Let }}\,\,\,\,\,u = {\tan ^{ - 1}}x,\,\,\,\,\,du = \frac{1}{{1 + {x^2}}}dx \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dv = \frac{1}{{{x^2}}},\,\,\,\,\,v = - \frac{1}{x} \cr
& \int {udv} = uv - \int {vdu} \,\,\,\, \cr
& \to \,\,\,\int {\frac{{{{\tan }^{ - 1}}x}}{{{x^2}}}} dx = - \frac{1}{x}{\tan ^{ - 1}}x - \int {\left( { - \frac{1}{x}} \right)\left( {\frac{1}{{1 + {x^2}}}} \right)} dx \cr
& = - \frac{1}{x}{\tan ^{ - 1}}x + \int {\frac{1}{{x\left( {1 + {x^2}} \right)}}} dx \cr
& \cr
& {\text{Decomposing the integrand }}\frac{1}{{x\left( {1 + {x^2}} \right)}}{\text{ into partial fractions}} \cr
& \frac{1}{{x\left( {1 + {x^2}} \right)}} = \frac{A}{x} + \frac{{Bx + C}}{{{x^2} + 1}} \cr
& \,\,\,{\text{Multiply by }}x\left( {1 + {x^2}} \right) \cr
& \,\,\,1 = A\left( {{x^2} + 1} \right) + \left( {Bx + C} \right)x \cr
& \,\,\,1 = A{x^2} + A + B{x^2} + Cx \cr
& \,\,\,1 = \left( {A{x^2} + B{x^2}} \right) + Cx + A \cr
& \,\,{\text{Equating coefficients }}A + B = 0,\,\,\,\,C = 0,\,\,\,A = 1,\,\,\,\,B = - 1 \cr
& \,\,\,\,\frac{1}{{x\left( {1 + {x^2}} \right)}} = \frac{A}{x} + \frac{{Bx + C}}{{{x^2} + 1}} = \frac{1}{x} - \frac{x}{{{x^2} + 1}} \cr
& {\text{Then}}{\text{,}} \cr
& - \frac{1}{x}{\tan ^{ - 1}}x + \int {\frac{1}{{x\left( {1 + {x^2}} \right)}}} dx = - \frac{1}{x}{\tan ^{ - 1}}x + \int {\left( {\frac{1}{x} - \frac{x}{{{x^2} + 1}}} \right)} dx \cr
& \cr
& {\text{Integrate}} \cr
& = - \frac{1}{x}{\tan ^{ - 1}}x + \ln \left| x \right| - \frac{1}{2}\ln \left( {{x^2} + 1} \right) + C \cr} $$
