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 $z$ |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $e^{a x}$: $14.517$