Primitive of Inverse Hyperbolic Tangent of x over a over x squared

Theorem

 * $\ds \int \frac 1 {x^2} \artanh \dfrac x a \rd x = -\frac 1 x \artanh \dfrac x a + \frac 1 {2 a} \map \ln {\frac {x^2} {a^2 - x^2} } + C$

Proof
With a view to expressing the primitive in the form:
 * $\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$

let:

and let:

Then:

Also see

 * Primitive of $\dfrac 1 {x^2} \arsinh \frac x a$


 * Primitive of $\dfrac 1 {x^2} \arcosh \frac x a$


 * Primitive of $\dfrac 1 {x^2} \arcoth \frac x a$