# Primitive of x squared by Arccotangent of x over a

## Theorem

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

## 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:

 $\displaystyle u$ $=$ $\displaystyle \operatorname{arccot} \frac x a$ $\displaystyle \implies \ \$ $\displaystyle \frac {\mathrm d u} {\mathrm d x}$ $=$ $\displaystyle \frac {-a} {x^2 + a^2}$ Derivative of $\operatorname{arccot} \dfrac x a$

and let:

 $\displaystyle \frac {\mathrm d v} {\mathrm d x}$ $=$ $\displaystyle x^2$ $\displaystyle \implies \ \$ $\displaystyle v$ $=$ $\displaystyle \frac {x^3} 3$ Primitive of Power

Then:

 $\displaystyle \int x^2 \operatorname{arccot} \frac x a \ \mathrm d x$ $=$ $\displaystyle \frac {x^3} 3 \operatorname{arccot} \frac x a - \int \frac {x^3} 3 \left({\frac {-a} {x^2 + a^2} }\right) \ \mathrm d x + C$ Integration by Parts $\displaystyle$ $=$ $\displaystyle \frac {x^3} 3 \operatorname{arccot} \frac x a + \frac a 3 \int \frac {x^3 \ \mathrm d x} {x^2 + a^2} + C$ Primitive of Constant Multiple of Function $\displaystyle$ $=$ $\displaystyle \frac {x^3} 3 \operatorname{arccot} \frac x a + \frac a 3 \left({\frac {x^2} 2 - \frac {a^2} 2 \ln \left ({x^2 + a^2}\right) }\right) + C$ Primitive of $\dfrac {x^3} {x^2 + a^2}$ $\displaystyle$ $=$ $\displaystyle \frac {x^3} 3 \operatorname{arccot} \frac x a + \frac {a x^2} 6 - \frac {a^3} 6 \ln \left({x^2 + a^2}\right) + C$ simplifying

$\blacksquare$