Evaluate integral with quadratic expression without root in the denominator $$\int \frac{1}{x(x^2+1)}dx = ? $$ 
How to solve it? Expanding to $\frac {A}{x}+ \frac{Bx +C}{x^2+1}$ would be wearisome.

And just to make this problem unnecessarily complicated:
$$I=\int\frac{dx}{x^2\left(x+\frac{1}{x}\right)}=\int\frac{dx}{\left(x+\frac{1}{x}\right)}-\int\frac{\left(1-\frac{1}{x^2}\right)dx}{\left(x+\frac{1}{x}\right)}=\frac{1}{2}\int\frac{d\left(x^2+1\right)}{x^2+1}-\int\frac{d\left(x+\frac{1}{x}\right)}{\left(x+\frac{1}{x}\right)}$$
