Primitive of Power of x by Arctangent of x over a/Proof 2

Theorem

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

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:

$\ds u$

$=$

$\ds \arctan \frac x a$

$\ds \leadsto \ \ $

$\ds \frac {\d u} {\d x}$

$=$

$\ds \frac a {x^2 + a^2}$

and let:

$\ds \frac {\d v} {\d x}$

$=$

$\ds x^m$

$\ds \leadsto \ \ $

$\ds v$

$=$

$\ds \frac {x^{m + 1} } {m + 1}$

Then:

$\ds \int x^m \arctan \frac x a \rd x$

$=$

$\ds \frac {x^{m + 1} } {m + 1} \arctan \frac x a - \int \frac {x^{m + 1} } {m + 1} \paren {\frac a {x^2 + a^2} } \rd x$

$\ds $

$=$

$\ds \frac {x^{m + 1} } {m + 1} \arctan \frac x a - \frac a {m + 1} \int \frac {x^{m + 1} \rd x} {x^2 + a^2}$

$\blacksquare$