Primitive of Arctangent of a x

Theorem

$\ds \int \arctan a x \rd x = x \arctan a x - \dfrac 1 {2 a} \map \ln {1 + a^2 x^2} + C$

$\ds \int \arctan x \rd x$

$=$

$\ds x \arctan x - \dfrac 1 2 \map \ln {1 + x^2} + C$

$\ds \leadsto \ \ $

$\ds \int \arctan a x \rd x$

$=$

$\ds \dfrac 1 a \paren {\paren {a x} \arctan a x - \dfrac 1 2 \map \ln {1 + a^2 x^2} } + C$

$\ds $

$=$

$\ds x \arctan a x - \dfrac 1 {2 a} \map \ln {1 + a^2 x^2} + C$

simplifying

$\blacksquare$