Answer
$$\ln x-\frac{1}{2} \ln \left|1+x^{2}\right|-\frac{\tan ^{-1} x}{x}+C$$
Work Step by Step
Given $$\int x^{-2} \tan ^{-1} x d x$$
Let
\begin{align*}
u&= \tan ^{-1} x\ \ \ \ \ \ dv=x^{-2} d x\\
du&=\frac{1}{1+x^2} dx\ \ \ \ \ \ v=-x^{-1}
\end{align*}
Then
\begin{align*}
\int x^{-2} \tan ^{-1} x d x&= -\frac{\tan ^{-1} x}{x}+\int \frac{1}{x\left(1+x^{2}\right)} d x
\end{align*}
Since
\begin{aligned}
\frac{1}{x\left(1+x^{2}\right)} &=\frac{A}{x}+\frac{B x+C}{1+x^{2}} \\
1 &=A\left(1+x^{2}\right)+(B x+C) x
\end{aligned}
At $ x=0 $, $A= 1$ and by comparing $ C= 0,\ \ B= -1$, we have:
\begin{aligned}
\int x^{-2} \tan ^{-1} x d x &=-\frac{\tan ^{-1} x}{x}+\int \frac{1}{x\left(1+x^{2}\right)} d x \\
&=-\frac{\tan ^{-1} x}{x}+\int \frac{1}{x} d x-\int \frac{x}{\left(1+x^{2}\right)} d x \\
&=\ln x-\frac{1}{2} \ln \left|1+x^{2}\right|-\frac{\tan ^{-1} x}{x}+C
\end{aligned}
