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

Theorem

 * $\ds \int x^2 \artanh \frac x a \rd x = \frac {a x^2} 6 + \frac {x^3} 3 \artanh \frac x a + \frac {a^3} 6 \map \ln {a^2 - x^2} + C$

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 $x^2 \arsinh \dfrac x a$


 * Primitive of $x^2 \arcosh \dfrac x a$


 * Primitive of $x^2 \arcoth \dfrac x a$