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

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

Then:
 * $\ds \int \frac {x \rd x} {a x^2 + b x + c} = \frac 1 {2 a} \ln \size {a x^2 + b x + c} - \frac b {2 a} \int \frac {\d x} {a x^2 + b x + c}$

Proof
First note that by Derivative of Power:
 * $(1): \quad \map {\dfrac \d {\d x} } {a x^2 + b x + c} = 2 a x + b$

Then:

Also see

 * Primitive of $\dfrac 1 {a x^2 + b x + c}$