Primitive of Reciprocal of p by Exponential of a x plus q by Exponential of -a x

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Theorem

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


Proof

\(\ds z\) \(=\) \(\ds e^{a x}\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d z} {\d x}\) \(=\) \(\ds a e^{a x}\) Derivative of $e^{a x}$
\(\ds \leadsto \ \ \) \(\ds \int \frac {\d x} {p e^{a x} + q e^{-a x} }\) \(=\) \(\ds \int \frac 1 {p z + q z^{-1} } \frac {\d z} {a z}\) Integration by Substitution
\(\ds \) \(=\) \(\ds \frac 1 {a p} \int \frac {\d z} {z^2 + \dfrac q p}\) Primitive of Constant Multiple of Function


Let $\dfrac q p > 0$.

Then:

\(\ds \frac 1 {a p} \int \frac {\d z} {z^2 + \dfrac q p}\) \(=\) \(\ds \frac 1 {a p} \paren {\frac 1 {\sqrt {q / p} } \arctan \frac z {\sqrt {q / p} } } + C\) Primitive of $\dfrac 1 {x^2 + a^2}$
\(\ds \) \(=\) \(\ds \frac 1 {a \sqrt {p q} } \map \arctan {\sqrt {\frac p q} z}\) simplifying
\(\ds \) \(=\) \(\ds \frac 1 {a \sqrt {p q} } \map \arctan {\sqrt {\frac p q} e^{a x} }\) substituting for $z$


Let $\dfrac q p < 0$.

Then let $d^2 = -\dfrac q p$ and so:

\(\ds \frac 1 {a p} \int \frac {\d z} {z^2 + \dfrac q p}\) \(=\) \(\ds \frac 1 {a p} \int \frac {\d z} {z^2 - d^2}\)
\(\ds \) \(=\) \(\ds \frac 1 {a p} \paren {\frac 1 {2 d} \ln \size {\frac {z - d} {z + d} } } + C\) Primitive of $\dfrac 1 {x^2 - a^2}$
\(\ds \) \(=\) \(\ds \frac 1 {a p} \paren {\frac 1 {2 \sqrt {-q / p} } \ln \size {\frac {z - \sqrt {-q / p} } {z + \sqrt {-q / p} } } } + C\) substituting for $d$
\(\ds \) \(=\) \(\ds \frac 1 {2 a \sqrt {-p q} } \ln \size {\frac {z - \sqrt {-\frac q p} } {z + \sqrt {-\frac q p} } } + C\) simplifying
\(\ds \) \(=\) \(\ds \frac 1 {2 a \sqrt {-p q} } \ln \size {\frac {e^{a x} - \sqrt {-\frac q p} } {e^{a x} + \sqrt {-\frac q p} } } + C\) substituting for $z$

$\blacksquare$


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