Primitive of x squared over x squared plus a squared

Theorem

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

where $a$ is a non-zero constant.

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

$=$

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

long division

$\ds $

$=$

$\ds \int \d x - a^2 \int \frac {\d x} {x^2 + a^2}$

$\ds $

$=$

$\ds x - a^2 \int \frac {\d x} {x^2 + a^2} + C$

$\ds $

$=$

$\ds x - a^2 \frac 1 a \arctan {\frac x a} + C$

$\ds $

$=$

$\ds x - a \arctan {\frac x a} + C$

simplifying

$\blacksquare$