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

Is there an alternative

Answer: 
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

$$\boxed{\displaystyle\int \frac{\tan^{-1} x}{x^2} dx = -\frac 1 2 \ln(1 + x^2) + \ln x - \frac{\tan^{-1} x}{x} + C}$$
$$\int\frac{\arctan(x)}{x^{2}}dx=x\frac{\arctan(x)}{x^{2}}-\int x\frac{\frac{x^{2}}{1+x^{2}}-2x\arctan(x)}{x^{4}}dx$$

$$=\frac{\arctan(x)}{x}-\int\frac{1}{x(1+x^{2})}dx+2\int\frac{\arctan(x)}{x^{2}}dx$$

$$=\frac{\arctan(x)}{x}-\int\frac{1}{x}dx+\frac{1}{2}\int\frac{x}{1+x^{2}}dx+2\int\frac{\arctan(x)}{x^{2}}dx$$

So

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