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

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

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

Proof
First:

Then: