Primitive of Reciprocal of p plus q by Hyperbolic Cotangent of a x

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Theorem

$\ds \int \frac {\d x} {p + q \coth a x} = \frac {p x} {p^2 - q^2} - \frac q {a \paren {p^2 - q^2} } \ln \size {p \sinh a x + q \cosh a x} + C$


Proof

We have:

$\dfrac \d {\d x} \paren {p \sinh a x + q \cosh a x} = a p \cosh a x + a q \sinh a x$

Thus:

\(\ds \int \frac {\d x} {p + q \coth a x}\) \(=\) \(\ds \int \frac {\d x} {p + q \dfrac {\cosh a x} {\sinh a x} }\) Definition 2 of Hyperbolic Cotangent
\(\ds \) \(=\) \(\ds \int \frac {\sinh a x \rd x} {p \sinh a x + q \cosh a x}\)
\(\ds \) \(=\) \(\ds \frac 1 {p^2 - q^2} \int \frac {\paren {p^2 - q^2} \sinh a x \rd x} {p \sinh a x + q \cosh a x}\)
\(\ds \) \(=\) \(\ds \frac 1 {p^2 - q^2} \int \frac {p^2 \sinh a x + p q \cosh a x - p q \cosh a x - q^2 \sinh a x} {p \sinh a x + q \cosh a x} \rd x\)
\(\ds \) \(=\) \(\ds \frac 1 {p^2 - q^2} \paren {\int \frac {p^2 \sinh a x + p q \cosh a x} {p \sinh a x + q \cosh a x} \rd x - \int \frac {p q \cosh a x + q^2 \sinh a x} {p \sinh a x + q \cosh a x} \rd x}\)
\(\ds \) \(=\) \(\ds \frac 1 {p^2 - q^2} \paren {\int p \rd x - \frac q a \int \frac {\map \d {p \sinh a x + q \cosh a x} } {p \sinh a x + q \cosh a x} }\)
\(\ds \) \(=\) \(\ds \frac {p x} {p^2 - q^2} - \frac q {a \paren {p^2 - q^2} } \ln \size {p \sinh a x + q \cosh a x} + C\) Primitive of Constant and Primitive of Reciprocal

$\blacksquare$


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