Primitive of x squared by Arctangent of x over a

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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:

\(\displaystyle u\) \(=\) \(\displaystyle \arctan \frac x a\)
\(\displaystyle \implies \ \ \) \(\displaystyle \frac {\mathrm d u} {\mathrm d x}\) \(=\) \(\displaystyle \frac a {x^2 + a^2}\) Derivative of $\arctan \dfrac x a$


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:

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

$\blacksquare$


Also see


Sources