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.

*Just do it!*$\,$: break the integrand into partial fractions...
Using partial fractions is the most straightforward approach (and in this case, is hardly wearisome).
$$ \frac{1}{x(x^2 + 1)} = \frac{x^2+1-x^2}{x(x^2+1)} = \frac{\color{blue} {\bf{x^2+1}}}{x\color{blue}{\bf(x^2+1)}} - \frac{\color{blue}{\bf x}\cdot x}{\color{blue}{\bf{x}}(x^2+1)}=\frac1x - \frac{x}{x^2+1}$$

$$\text{So now you have}\;\;\;\int \dfrac{dx}{x(x^2+1)}\; = \;\int \dfrac{1}x\,dx - \int\dfrac{x}{x^2+1}\,dx$$
and I trust you can take it from here.
The more you work with partial fractions (i.e. practice, i.e.), the easier they are to work with, and the more quickly you'll see how to proceed in doing so. 
