$\int \dfrac{dx}{x^2 - a^2} $

The correct answer is $\dfrac{1}{2a} log \left| \dfrac{x - a}{x + a} \right|$ + $C$

$\int \dfrac{dx}{x^2 - a^2} $

$\int \dfrac{dx}{(x - a)(x + a)} $

$\int \dfrac{A}{(x - a)} . dx + \int \dfrac{B}{(x + a)} . dx$ ---- (i)


Solving for values of A and B, we get

$1 = A(x + a) + B(x - a)$

Putting $x = a$, we get $A = \dfrac{1}{2a}$

Putting $x = -a$, we get $A = -\dfrac{1}{2a}$


Substituting these values in (i),

$\int \dfrac{1}{2a(x - a)} . dx - \int \dfrac{1}{2a(x + a)} . dx$

$\dfrac{1}{2a} \left[ \int \dfrac{1}{(x - a)} . dx - \int \dfrac{1}{(x + a)} . dx \right]$

$\dfrac{1}{2a} \left[ log{|x - a|} - log{|x + a|} \right]$ + $C$

$\dfrac{1}{2a} log \left| \dfrac{x - a}{x + a} \right|$ + $C$