Answer
$$\int \frac{\tan ^{-1} x}{x^{2}} d x=-\frac{1}{x} \tan ^{-1} x+\ln |x|-\frac{1}{2} \ln \left(1+x^{2}\right)+C$$
Work Step by Step
Use the formula $$\int x^{n} \tan ^{-1} a x d x=\frac{x^{n+1}}{n+1} \tan ^{-1} a x-\frac{a}{n+1} \int \frac{x^{n+1} d x}{1+a^{2} x^{2}}, \quad n \neq-1$$
with $n=-2,\ a=1$
\begin{align*}
\int \frac{\tan ^{-1} x}{x^{2}} d x&=\int x^{-2} \tan ^{-1} x d x\\
&=\frac{x^{(-2+1)}}{(-2+1)} \tan ^{-1} x-\frac{1}{(-2+1)} \int \frac{x^{(-2+1)}}{1+x^{2}} d x\\
&=\frac{x^{-1}}{(-1)} \tan ^{-1} x+\int \frac{x^{-1}}{1+x^{2}} d x
\end{align*}
and
\begin{align*}\int \frac{x^{-1} d x}{1+x^{2}}&=\int \frac{d x}{x\left(1+x^{2}\right)}\\
&=\int \frac{d x}{x}-\int \frac{x d x}{1+x^{2}}\\
&=\ln |x|-\frac{1}{2} \ln \left(1+x^{2}\right)+C
\end{align*}
Hence
$$\int \frac{\tan ^{-1} x}{x^{2}} d x=-\frac{1}{x} \tan ^{-1} x+\ln |x|-\frac{1}{2} \ln \left(1+x^{2}\right)+C$$
