Primitive of Reciprocal of a x squared plus b x plus c/c equal to 0/Proof 2

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

Let $c = 0$.

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

Proof
Let $c = 0$.

From Primitive of Reciprocal of a x squared plus b x plus c, we have:
 * $\displaystyle \int \frac {\mathrm d x} {a x^2 + b x + c} = \begin{cases}

\dfrac 2 {\sqrt {4 a c - b^2} } \arctan \left({\dfrac {2 a x + b} {\sqrt {4 a c - b^2} } }\right) + C & : b^2 - 4 a c < 0 \\ \dfrac 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 & : b^2 - 4 a c > 0 \\ \dfrac {-2} {2 a x + b} + C & : b^2 = 4 a c \end{cases}$

As $b \ne 0$ it follows that $b^2 - 4 a c = b^2 > 0$.

Thus: