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

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Theorem

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

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} } = \begin {cases}

\dfrac 1 {\sqrt a} \ln \size {2 \sqrt a \sqrt {a x^2 + b x + c} + 2 a x + b} + C & : b^2 - 4 a c > 0 \\ \dfrac 1 {\sqrt a} \map \arsinh {\dfrac {2 a x + b} {\sqrt {4 a c - b^2} } } + C & : b^2 - 4 a c < 0 \\ \dfrac 1 {\sqrt a} \ln \size {2 a x + b} + C & : b^2 - 4 a c = 0 \end {cases}$ where $\arsinh$ denotes the area hyperbolic sine function.


Proof

Completing the Square

First:

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


Put:

\(\ds z\) \(=\) \(\ds 2 a x + b\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d z} {\d x}\) \(=\) \(\ds 2 a\) Derivative of Power


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

Thus:

\(\ds \int \frac {\d x} {\sqrt {a x^2 + b x + c} }\) \(=\) \(\ds \int \frac {2 \sqrt a \rd x} {\sqrt {\paren {2 a x + b}^2 - \paren {b^2 - 4 a c} } }\) from $(1)$
\(\ds \) \(=\) \(\ds \int \frac {\d z} {\sqrt a \sqrt {z^2 - D} }\) Integration by Substitution
\(\ds \) \(=\) \(\ds \frac 1 {\sqrt a} \int \frac {\d z} {\sqrt {z^2 - D} }\) Primitive of Constant Multiple of Function


Positive Discriminant

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

Then:

\(\ds D\) \(>\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds D\) \(=\) \(\ds q^2\) for some $q \in \R$
\(\ds \leadsto \ \ \) \(\ds q\) \(=\) \(\ds \sqrt {b^2 - 4 a c}\) by definition of $D$


Thus:

\(\ds \int \frac {\d x} {\sqrt {a x^2 + b x + c} }\) \(=\) \(\ds \frac 1 {\sqrt a} \int \frac {\d z} {\sqrt {z^2 - q^2} }\) Integration by Substitution
\(\ds \) \(=\) \(\ds \frac 1 {\sqrt a} \ln \size {z + \sqrt {z^2 - q^2} } + C\) Primitive of $\dfrac 1 {\sqrt {x^2 - a^2} }$
\(\ds \) \(=\) \(\ds \frac 1 {\sqrt a} \ln \size {2 a x + b + \sqrt {\paren {2 a x + b}^2 - \paren {b^2 - 4 a c} } } + C\) substituting for $z$ and $q$
\(\ds \) \(=\) \(\ds \frac 1 {\sqrt a} \ln \size {2 a x + b + \sqrt {4 a \paren {a x^2 + b x + c} } } + C\) Completing the Square
\(\ds \) \(=\) \(\ds \frac 1 {\sqrt a} \ln \size {2 \sqrt a \sqrt {a x^2 + b x + c} + 2 a x + b} + C\) rearranging


$\Box$


Negative Discriminant

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

Then:

\(\ds - D\) \(>\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds -D\) \(=\) \(\ds q^2\) for some $q \in \R$
\(\ds \leadsto \ \ \) \(\ds q\) \(=\) \(\ds \sqrt {4 a c - b^2}\) by definition of $D$


Thus:

\(\ds \int \frac {\d x} {\sqrt{a x^2 + b x + c} }\) \(=\) \(\ds \frac 1 {\sqrt a} \int \frac {\d z} {\sqrt {z^2 + q^2} }\) Integration by Substitution
\(\ds \) \(=\) \(\ds \frac 1 {\sqrt a} \map \arsinh {\frac z q} + C\) Primitive of $\dfrac 1 {\sqrt {x^2 + a^2} }$
\(\ds \) \(=\) \(\ds \frac 1 {\sqrt a} \map \arsinh {\frac {2 a x + b} {\sqrt {4 a c - b^2} } } + C\) substituting for $z$ and $q$


$\Box$


Zero Discriminant

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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