Primitive of x over square of a x squared plus b x plus c

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

Then:
 * $\ds \int \frac {x \rd x} {\paren {a x^2 + b x + c}^2} = \frac {-\paren {b x + 2 c} } {\paren {4 a c - b^2} \paren {a x^2 + b x + c} } - \frac b {4 a c - b^2} \int \frac {\d x} {a x^2 + b x + c}$

Proof
Let:

Then: