Integral with partial fractions I'm trying to calculate the following integral:

$$I=\int \frac{\arctan(x)}{x^4}dx$$

My steps so far are:
Per partes:
$$\frac{-\arctan(x)}{3x^3}+\int{\frac{1}{1+x^2} \frac{1}{3x^3}}dx
=\frac{-\arctan(x)}{3x^3}+\frac{1}{3}\int{\frac{1}{1+x^2} \frac{1}{x^3}}dx$$
and now I want to do partial fractions. However, with this integral, I fail to do partial fractions. Could you help me?
Thanks

Or more simply:
$$A=\frac{1}{1+x^2}*\frac{1}{x^3}$$
$$A=-\frac 1 x \left (\frac{1}{1+x^2}-\frac{x}{x^3} \right )$$
$$A=-\frac{1}{x(1+x^2)}+\frac{1}{x^3}$$

$$B=\frac{1}{x(1+x^2)}=\frac{1}{x}-\frac x {(1+x^2)}$$
We have that:
$$I=\int \left \{\frac x {(1+x^2)}-\frac{1}{x}+\frac{1}{x^3}  \right \}dx$$
