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

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Theorem

Let $c = 0$.

Then:

$\ds \int \frac {\d x} {a x^2 + b x + c} = \frac 1 b \ln \size {\frac x {a x + b} } + C$


Proof

Let $c = 0$.

From Primitive of $\dfrac 1 {a x^2 + b x + c}$, we have:

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

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


Thus:

\(\ds \int \frac {\d x} {a x^2 + b x + c}\) \(=\) \(\ds \frac 1 {\sqrt {b^2 - 0} } \ln \size {\dfrac {2 a x + b - \sqrt {b^2 - 0} } {2 a x + b + \sqrt {b^2 - 0} } } + C\)
\(\ds \) \(=\) \(\ds \frac 1 {\sqrt {b^2} } \ln \size {\frac {2 a x + b - b} {2 a x + b + b} } + C\) simplifying
\(\ds \) \(=\) \(\ds \frac 1 b \ln \size {\frac {a x} {a x + b} } + C\) further simplifying
\(\ds \) \(=\) \(\ds \frac 1 b \paren {\ln \size {\frac x {a x + b} } + \ln \size a} + C\) simplifying
\(\ds \) \(=\) \(\ds \frac 1 b \ln \size {\frac x {a x + b} } + C\) subsuming $\dfrac 1 b \ln \size a$ into the arbitrary constant

$\blacksquare$