Primitive of Reciprocal of x squared minus a squared/Logarithm Form 2/Proof 2

Theorem

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

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

$=$

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

$\ds $

$=$

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

$\ds $

$=$

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

$\ds $

$=$

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

$\ds $

$=$

$\ds \dfrac 1 {2 a} \ln \size {\dfrac {x - a} {x + a} } + C$

$\blacksquare$