what is a smart way to find $\int \frac{\arctan\left(x\right)}{x^{2}}\,{\rm d}x$ I tried integration by parts, which gets very lengthy due to partial fractions.
Is there an alternative

It is not so lenghty. If $$I =\int \frac{\tan^{-1}x}{x^2}\,dx$$ let start with $u=\tan^{-1}x$ and $v'=\frac{\tan^{-1}x}{x^2}\,dx$. So $u'=\frac{\text{dx}}{x^2+1}$ and $v=-\frac{1}{x}$ and then $$I =\int \frac{\tan^{-1}x}{x^2}\,dx=-\frac{\tan ^{-1}(x)}{x}+\int \frac{dx}{x \left(x^2+1\right)}$$ Now, using partial fraction decomposition $$\frac{1}{x \left(x^2+1\right)}=\frac{1}{x}-\frac{x}{x^2+1}$$ and, finally $$I =\int \frac{\tan^{-1}x}{x^2}\,dx=\log (x)-\frac{1}{2} \log \left(x^2+1\right)-\frac{\tan ^{-1}(x)}{x}+C$$
Added later after André Nicolas comment
If you really want to avoid partial fraction decomposition, use the substitution $x=\frac{1}{y}$; this gives $$I =\int \frac{\tan^{-1}x}{x^2}\,dx=-\int \tan ^{-1}\left(\frac{1}{y}\right)\,dy $$ and a single integration by parts will give $$I=-\frac{1}{2} \log \left(y^2+1\right)-y \tan ^{-1}\left(\frac{1}{y}\right)+C$$
