Find the integral $\int \frac{1}{x^2 \cdot \tan(x)} \ dx$ This problem seems pretty tricky. I need to find the integral of
$$\int \dfrac{1}{x^2 \cdot \tan(x)} \ dx$$
Any help would be greatly appreciated!

I would use the decomposition of $\frac{1}{\tan x}$ into simple fractions:  
$$\frac{1}{\tan x}=\frac{1}{x}+\sum_{n=1}^{\infty}\frac{2x}{x^2-n^2\pi^2}$$ So  
$$\frac{1}{x^2\tan x}=\frac{1}{x^3}+2\sum_{n=1}^{\infty}\frac{1}{x(x^2-n^2\pi^2)}$$ and  
$$\int\frac{dx}{x^2\tan x}=-\frac{1}{2x^2}+\sum_{n=1}^{\infty}\frac{\ln\frac{\left |x^2-n^2\pi^2  \right |}{x^2}}{n^2\pi^2}+\text{const}$$  
This is of course a matter of taste, whether the form is closed or not.
