Primitive of x by Square of Hyperbolic Cotangent of a x

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Theorem

$\displaystyle \int x \coth^2 a x \ \mathrm d x = \frac {x^2} 2 - \frac {x \coth a x} a + \frac 1 {a^2} \ln \left\vert{\sinh a x}\right\vert + C$


Proof

With a view to expressing the primitive in the form:

$\displaystyle \int u \frac {\mathrm d v}{\mathrm d x} \ \mathrm d x = u v - \int v \frac {\mathrm d u}{\mathrm d x} \ \mathrm d x$

let:

\(\displaystyle u\) \(=\) \(\displaystyle x\)
\(\displaystyle \implies \ \ \) \(\displaystyle \frac {\mathrm d u}{\mathrm d x}\) \(=\) \(\displaystyle 1\) Derivative of Identity Function


and let:

\(\displaystyle \frac {\mathrm d v}{\mathrm d x}\) \(=\) \(\displaystyle \coth^2 a x\)
\(\displaystyle \implies \ \ \) \(\displaystyle v\) \(=\) \(\displaystyle x - \frac {\coth a x} a\) Primitive of $\coth^2 a x$


Then:

\(\displaystyle \int x \coth^2 a x \ \mathrm d x\) \(=\) \(\displaystyle x \left({x - \frac {\coth a x} a}\right) - \int \left({x - \frac {\coth a x} a}\right) \times 1 \ \mathrm d x + C\) Integration by Parts
\(\displaystyle \) \(=\) \(\displaystyle x^2 - \frac {x \coth a x} a + \int x \ \mathrm d x + \frac 1 a \int \coth a x \ \mathrm d x + C\) Linear Combination of Integrals
\(\displaystyle \) \(=\) \(\displaystyle x^2 - \frac {x \coth a x} a + \frac {x^2} 2 + \frac 1 a \int \coth a x \ \mathrm d x + C\) Primitive of Power
\(\displaystyle \) \(=\) \(\displaystyle x^2 - \frac {x \coth a x} a + \frac {x^2} 2 + \frac 1 a \frac {\ln \left\vert{\sinh a x}\right\vert} a + C\) Primitive of $\coth a x$
\(\displaystyle \) \(=\) \(\displaystyle \frac {x^2} 2 - \frac {x \coth a x} a + \frac 1 {a^2} \ln \left\vert{\sinh a x}\right\vert + C\) simplifying

$\blacksquare$

Also see


Sources