Primitive of Reciprocal of a x squared plus b x plus c/c equal to 0/Proof 1
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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
First:
\(\ds c\) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \int \frac {\d x} {a x^2 + b x + c}\) | \(=\) | \(\ds \int \frac {\d x} {a x^2 + b x}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \int \frac {\d x} {x \paren {a x + b} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 b \ln \size {\frac x {a x + b} } + C\) | Primitive of $\dfrac 1 {x \paren {a x + b} }$ |
$\blacksquare$