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

Is there an alternative

Answer: 
Put $\tan^{ -1}x = y$ 

Then it becomes $\tan y$ = $x$. 

Then differentiate w.r.t $y$ then $dx = \sec^2y dy$.

Then finally,

$$I = \int y \csc^2 y dy$$ 

Then finally apply integration by parts. It will be little easier than directly applying Integration by parts.
$$\int\frac{\arctan(x)}{x^{2}}dx=x\frac{\arctan(x)}{x^{2}}-\int x\frac{\frac{x^{2}}{1+x^{2}}-2x\arctan(x)}{x^{4}}dx$$

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

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

So

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