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

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

Let $b = 0$.

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

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

Proof
Let $b = 0$.

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

\dfrac 2 {\sqrt {4 a c - b^2} } \map \arctan {\dfrac {2 a x + b} {\sqrt {4 a c - b^2} } } + C & : b^2 - 4 a c < 0 \\ \dfrac 1 {\sqrt {b^2 - 4 a c} } \ln \size {\dfrac {2 a x + b - \sqrt {b^2 - 4 a c} } {2 a x + b + \sqrt {b^2 - 4 a c} } } + C & : b^2 - 4 a c > 0 \\ \dfrac {-2} {2 a x + b} + C & : b^2 = 4 a c \end{cases}$

Let $a c > 0$.

Then $b^2 - 4 a c = 0 - 4 a c < 0$ and so:

Let $a c < 0$.

Then $b^2 - 4 a c = 0 - 4 a c > 0$ and so:

Let $c = 0$.

Then $b^2 - 4 a c = 0 - 0 = 0$ and so: