Primitive of x squared by Arctangent of x over a

Theorem

 * $\displaystyle \int x^2 \arctan \frac x a \ \mathrm d x = \frac {x^3} 3 \arctan \frac x a - \frac {a x^2} 6 + \frac {a^3} 6 \ln \left({x^2 + a^2}\right) + C$

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:

and let:

Then:

Also see

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


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


 * Primitive of $x^2 \operatorname{arccot} \dfrac x a$


 * Primitive of $x^2 \operatorname{arcsec} \dfrac x a$


 * Primitive of $x^2 \operatorname{arccsc} \dfrac x a$