Primitive of Reciprocal of a x squared plus b x plus c/b equal to 0/Proof 1

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

Let $b = 0$.

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

\dfrac 1 {\sqrt {a c} } \arctan \left({x \sqrt {\dfrac a c} }\right) + C & : a c > 0 \\ \dfrac 1 {2 \sqrt {-a c} } \ln \left({\dfrac {a x - \sqrt {-a c} } {a x + \sqrt {-a c} } }\right) + C & : a c < 0 \\ \dfrac {-1} {a x} + C & : c = 0 \end{cases}$

Proof
First:

Let $a c > 0$.

Then $\dfrac c a > 0$ and:

Thus:

Let $a c < 0$.

Then $\dfrac c a > 0$ and:

Thus:

Let $c = 0$.

Then: