Primitive of Reciprocal of x squared by x squared plus a squared

Theorem

$\ds \int \frac {\d x} {x^2 \paren {x^2 + a^2} } = -\frac 1 {a^2 x} - \frac 1 {a^3} \arctan \frac x a + C$

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

$=$

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

$\ds $

$=$

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

$\ds $

$=$

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

$\ds $

$=$

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

$\ds $

$=$

$\ds -\frac 1 {a^2 x} - \frac 1 {a^3} \arctan \frac x a + C$

simplifying

$\blacksquare$