Primitive of Reciprocal of a x squared plus b x plus c/Positive Discriminant

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

Let $b^2 - 4 a c > 0$.

Then:


 * $\displaystyle \int \frac {\mathrm d x} {a x^2 + b x + c} = \frac 1 {\sqrt {b^2 - 4 a c} } \ln \left\vert{\dfrac {2 a x + b - \sqrt {b^2 - 4 a c} } {2 a x + b + \sqrt {b^2 - 4 a c} } }\right\vert + C$

Proof
First:

Put: $z = 2 a x + b$

Let $D = b^2 - 4 a c$.

Thus:

Let $b^2 - 4 a c > 0$.

Then:

Thus: