Answer
The given improper integral is:
$$
\int_{1}^{\infty} \frac{\tan ^{-1} x}{x^{2}} d x =\frac{\pi}{4}+\frac{1}{2} \ln 2.
$$
Work Step by Step
$$
\int_{1}^{\infty} \frac{\tan ^{-1} x}{x^{2}} d x
$$
First, we integrate the Indefinite Integral by parts
$$
\int \frac{\tan ^{-1} x}{x^{2}} d x
$$
with
$$
\left[\begin{array}{c}{u=\tan ^{-1} x , \quad\quad dv= \frac{dx}{x^{2}}} \\ {d u= \frac{d x}{1+x^{2} }, \quad\quad v= -\frac{1}{x} }\end{array}\right]
$$
$$
\begin{split}
\int \frac{\tan ^{-1} x}{x^{2}} d x &=\frac{-\tan ^{-1} x}{x}+\int \frac{1}{x} \frac{d x}{1+x^{2}} \\
& =\frac{-\tan ^{-1} x}{x}+\int\left[\frac{1}{x}-\frac{x}{x^{2}+1}\right] d x \\ &=\frac{-\tan ^{-1} x}{x}+\ln |x|-\frac{1}{2} \ln \left(x^{2}+1\right)+C
\\
&=\frac{-\tan ^{-1} x}{x}+\frac{1}{2} \ln \frac{x^{2}}{x^{2}+1}+C
\end{split}
$$
Now, we evaluate the given integral
$$
\int_{1}^{\infty} \frac{\tan ^{-1} x}{x^{2}} d x
$$
as follows:
$$
\begin{aligned} \int_{1}^{\infty} \frac{\tan ^{-1} x}{x^{2}} d x &=\lim _{t \rightarrow \infty}\left[-\frac{\tan ^{-1} x}{x}+\frac{1}{2} \ln \frac{x^{2}}{x^{2}+1}\right]_{1}^{t}\\
&=\lim _{t \rightarrow \infty}\left[-\frac{\tan ^{-1} t}{t}+\frac{1}{2} \ln \frac{t^{2}}{t^{2}+1}+\frac{\pi}{4}-\frac{1}{2} \ln \frac{1}{2}\right] \\ &=0+\frac{1}{2} \ln 1+\frac{\pi}{4}+\frac{1}{2} \ln 2\\
&=\frac{\pi}{4}+\frac{1}{2} \ln 2 \end{aligned}
$$
Thus the given improper integral is:
$$
\int_{1}^{\infty} \frac{\tan ^{-1} x}{x^{2}} d x =\frac{\pi}{4}+\frac{1}{2} \ln 2.
$$
