Question: I tried integration by parts, which gets very lengthy due to partial fractions.

Is there an alternative

Answer: 
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 $$I=-\frac{1}{2} \log \left(y^2+1\right)-y \tan ^{-1}\left(\frac{1}{y}\right)+C$$
Compute using the series expansion:

\begin{align\*}
\int \frac{\tan^{-1} x}{x^2} dx &= \int \frac 1 {x^2} \left(x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots\right) dx \\
&= \int \frac 1 x - \frac{x}{3} + \frac{x^3}{5} - \cdots dx \\
&= \ln x - \frac{x^2}{2 \cdot 3} + \frac{x^4}{4 \cdot 5} - \frac{x^6}{6 \cdot 7} + \cdots
\end{align\*}

Now using the identity

$$\frac 1 {n(n + 1)} = \frac 1 n - \frac 1 {n + 1}$$

we can rewrite the remaining series as

\begin{align\*}
\frac{x^2}{2 \cdot 3} - \frac{x^4}{4 \cdot 5} + \frac{x^6}{6 \cdot 7} - \cdots &= \left(\frac{x^2}{2} - \frac{x^4}{4} + \frac{x^6}{6} - \cdots\right) -\left(\frac{x^2}{3} - \frac{x^4}{5} + \frac{x^6}{7}\right)
\end{align\*}

These are both (related to) standard series for well-known functions; the second one is 

$$1-\frac {\tan^{-1} x} x$$

and the first is $\frac 1 2\ln(x^2 + 1)$, using the fact that $$\ln(1 + t) = t - \frac {t^2}2 + \frac{t^3}3 - \frac{t^4}4 + \cdots$$

Putting all this together gives the result

$$\boxed{\displaystyle\int \frac{\tan^{-1} x}{x^2} dx = -\frac 1 2 \ln(1 + x^2) + \ln x - \frac{\tan^{-1} x}{x} + C}$$