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

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

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

Then:


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

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: