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