# Primitive of Inverse Hyperbolic Tangent of x over a

## Theorem

$\displaystyle \int \tanh^{-1} \frac x a \rd x = x \tanh^{-1} \dfrac x a + \frac {a \ln \left({a^2 - x^2}\right)} 2 + C$

## Proof

With a view to expressing the primitive in the form:

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

let:

 $\displaystyle u$ $=$ $\displaystyle \tanh^{-1} \frac x a$ $\displaystyle \implies \ \$ $\displaystyle \frac {\d u} {\d x}$ $=$ $\displaystyle \frac a {a^2 - x^2}$ Derivative of $\tanh^{-1} \dfrac x a$

and let:

 $\displaystyle \frac {\d v} {\d x}$ $=$ $\displaystyle 1$ $\displaystyle \implies \ \$ $\displaystyle v$ $=$ $\displaystyle x$ Primitive of Constant

Then:

 $\displaystyle \int \tanh^{-1} \frac x a \rd x$ $=$ $\displaystyle x \tanh^{-1} \frac x a - \int x \left({\frac a {a^2 - x^2} }\right) \rd x + C$ Integration by Parts $\displaystyle$ $=$ $\displaystyle x \tanh^{-1} \frac x a - a \int \frac {x \rd x} {a^2 - x^2} + C$ Primitive of Constant Multiple of Function $\displaystyle$ $=$ $\displaystyle x \tanh^{-1} \frac x a - a \left({- \frac 1 2 \ln \left({a^2 - x^2}\right)}\right) + C$ Primitive of $\dfrac x {a^2 - x^2}$ $\displaystyle$ $=$ $\displaystyle x \tanh^{-1} \dfrac x a + \frac {a \ln \left({a^2 - x^2}\right)} 2 + C$ simplifying

$\blacksquare$