Primitive of x squared by Arccotangent of x over a

Theorem

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

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 $x^2 \arcsin \dfrac x a$


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


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


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


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