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

## 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 $z$

$\blacksquare$