Primitive of Reciprocal of x by x squared plus a squared/Proof 1

Theorem

$\ds \int \frac {\rd x} {x \paren {x^2 + a^2} } = \frac 1 {2 a^2} \map \ln {\frac {x^2} {x^2 + a^2} } + C$

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

$=$

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

$\ds $

$=$

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

$\ds $

$=$

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

$\ds $

$=$

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

$\ds $

$=$

$\ds \frac 1 {2 a^2} \map \ln {\frac {x^2} {x^2 + a^2} } + C$

$\blacksquare$