Primitive of Reciprocal of Root of a x squared plus b x plus c/a greater than 0/Zero Discriminant

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Theorem

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

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


Then for $x \in \R$ such that $a x^2 + b x + c > 0$:

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


Proof

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


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

Then:

\(\ds \int \frac {\d x} {\sqrt {a x^2 + b x + c} }\) \(=\) \(\ds \int \frac {2 \sqrt a \rd x} {\sqrt {\paren {2 a x + b}^2} }\) from $(1)$
\(\ds \) \(=\) \(\ds 2 \sqrt a \int \frac {\d x} {2 a x + b}\) Primitive of Constant Multiple of Function
\(\ds \) \(=\) \(\ds 2 \sqrt a \frac 1 {2 a} \ln \size {2 a x + b} + C\) Primitive of $\dfrac 1 {a x + b}$
\(\ds \) \(=\) \(\ds \frac 1 {\sqrt a} \ln \size {2 a x + b} + C\) simplifying

$\blacksquare$


Sources

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