Primitive of Power of x by Arctangent of a x

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Theorem

$\ds \int x^n \arctan a x \rd x = \frac {x^{n + 1} } {n + 1} \arctan a x + \frac a {n + 1} \int \frac {x^{n + 1} \rd x} {a^2 x^2 + 1}$

for $n \ne -1$.


Proof

Recall:

\(\text {(1)}: \quad\) \(\ds \int x^n \arctan x \rd x\) \(=\) \(\ds \frac {x^{n + 1} } {n + 1} \arctan x - \frac 1 {n + 1} \int \frac {x^{n + 1} \rd x} {x^2 + 1}\) Primitive of $x^n \arctan x$


Then:

\(\ds \int x^n \arctan a x \rd x\) \(=\) \(\ds \int \dfrac 1 {a^n} \paren {a x}^n \arctan a x \rd x\) manipulating into appropriate form
\(\ds \) \(=\) \(\ds \dfrac 1 {a^n} \int \paren {a x}^n \arctan a x \rd x\) Primitive of Constant Multiple of Function
\(\ds \) \(=\) \(\ds \dfrac 1 {a^n} \paren {\dfrac 1 a \paren {\frac {\paren {a x}^{n + 1} } {n + 1} \arctan a x + \frac 1 {n + 1} \int \frac {\paren {a x}^{n + 1} \rd x} {\paren {a x}^2 + 1} } }\) Primitive of Function of Constant Multiple, from $(1)$
\(\ds \) \(=\) \(\ds \frac {x^{n + 1} } {n + 1} \arctan a x + \frac a {n + 1} \int \frac {x^{n + 1} \rd x} {a^2 x^2 + 1}\) simplifying
There is believed to be a mistake here, possibly a typo.
In particular: can't resolve the $\dfrac a {n + 1}$ factor, it seems to want to be $\dfrac 1 {n + 1}$
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$\blacksquare$


Also see


Sources

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