Primitive of Reciprocal of x by square of a x squared plus b x plus c

Theorem
Let $a \in \R_{\ne 0}$.

Then:
 * $\displaystyle \int \frac {\d x} {x \paren {a x^2 + b x + c}^2} = \frac 1 {2 c \paren {a x^2 + b x + c} } - \frac b {2 c} \int \frac {\d x} {\paren {a x^2 + b x + c}^2} + \frac 1 c \int \frac {\d x} {x \paren {a x^2 + b x + c} }$