Answer
$$\displaystyle\int{\frac{\tan ^{-1}x}{x^2}dx}\\=\ln \left| \frac{x}{\sqrt{x^2+1}} \right|-\frac{\tan ^{-1}x}{x}+C$$
Work Step by Step
$
I=\displaystyle\int{\frac{\tan ^{-1}x}{x^2}dx}\\
I=\displaystyle\int{x^{-2}\tan ^{-1}x}\ dx\\
$
$\displaystyle{\left[\begin{matrix}
u=\tan ^{-1}x& du=\frac{1}{1+x^2}\\
dv=x^{-2}& v=-x^{-1}\\
\end{matrix}\right]}
$
$\displaystyle I=-\frac{\tan ^{-1}x}{x}-\int{-\frac{1}{x\left( 1+x^2 \right)}dx}\\
\displaystyle I=-\frac{\tan ^{-1}x}{x}+\int{\frac{1}{x\left( 1+x^2 \right)}dx}\\
\displaystyle I=-\frac{\tan ^{-1}x}{x}\color{orangered}{+\int{\frac{1}{x\left( 1+x^2 \right)}dx}}\\$
$
\displaystyle\color{orangered}{\int{\frac{1}{x\left( 1+x^2 \right)}dx}}$
$\displaystyle{\left[\begin{matrix}
x=\tan \theta& x^2=\tan ^2\theta& dx=\sec ^2\theta \ d\theta\\
\end{matrix}\right]}$
$\displaystyle\int{\frac{1}{\tan \theta \left( 1+\tan ^2\theta \right)}\sec ^2\theta \ d\theta}\\
\displaystyle\int{\frac{1}{\tan \theta \sec ^2\theta}\sec ^2\theta \,\,d\theta}\\
\displaystyle\int{\frac{1}{\tan \theta}\,\,d\theta}\\
\displaystyle\int{\cot \theta \ d\theta}\\
\displaystyle\color{orangered}{-\ln \left| \csc\theta \right|}$
$\displaystyle I=-\frac{\tan ^{-1}x}{x}\color{orangered}{-\ln \left| \csc\theta \right|}+C\\
\displaystyle I=-\frac{\tan ^{-1}x}{x}-\ln \left| \csc\theta \right|+C$
$\csc\theta =\frac{\sqrt{x^2+1}}{x}$
$\displaystyle I=-\frac{\tan ^{-1}x}{x}-\ln \left| \frac{\sqrt{x^2+1}}{x} \right|+C\\
\displaystyle I=\ln \left| \frac{x}{\sqrt{x^2+1}} \right|-\frac{\tan ^{-1}x}{x}+C$
