# Primitive of x by Square of Hyperbolic Cotangent of a x

## 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

$\blacksquare$