Primitive of Reciprocal of a x squared plus b x plus c/Zero Discriminant

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Theorem

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

Let $b^2 - 4 a c = 0$.

Then:

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


Proof

First:

\(\ds a x^2 + b x + c\) \(=\) \(\ds \frac {\paren {2 a x + b}^2 + 4 a c - b^2} {4 a}\) Completing the Square
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \int \frac {\d x} {a x^2 + b x + c}\) \(=\) \(\ds \int \frac {4 a \rd x} {\paren {2 a x + b}^2 - \paren {b^2 - 4 a c} }\)


Let $b^2 - 4 a c = 0$.

Then:

\(\ds \int \frac {\d x} {a x^2 + b x + c}\) \(=\) \(\ds \int \frac {4 a \rd x} {\paren {2 a x + b}^2}\) from $(1)$
\(\ds \) \(=\) \(\ds \frac {-4 a} {2 a \paren {2 a x + b} } + C\) Primitive of $\dfrac 1 {\paren {a x + b}^2}$
\(\ds \) \(=\) \(\ds \dfrac {-2} {2 a x + b} + C\) simplifying

$\blacksquare$


Sources

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