Answer
$$\frac{1}{{2a}}\ln \left| {\frac{{a - x}}{{a + x}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{{a^2} - {x^2}}}} dx \cr
& {\text{Decompose the integrand into partial practions}} \cr
& \frac{1}{{{a^2} - {x^2}}} = \frac{1}{{\left( {a + x} \right)\left( {a - x} \right)}} \cr
& \frac{1}{{\left( {a + x} \right)\left( {a - x} \right)}} = \frac{A}{{a + x}} + \frac{B}{{a - x}} \cr
& 1 = A\left( {a - x} \right) + B\left( {a + x} \right) \cr
& x = a \Rightarrow B = \frac{1}{{2a}} \cr
& x = - a \Rightarrow A = - \frac{1}{{2a}} \cr
& then \cr
& \frac{1}{{\left( {a + x} \right)\left( {a - x} \right)}} = - \frac{1}{{2a\left( {a + x} \right)}} + \frac{1}{{2a\left( {a - x} \right)}} \cr
& {\text{Rewrite}} \cr
& \int {\frac{1}{{{a^2} - {x^2}}}} dx = \int {\left( { - \frac{1}{{2a\left( {a + x} \right)}} + \frac{1}{{2a\left( {a - x} \right)}}} \right)} dx \cr
& {\text{Integrating}} \cr
& = - \frac{1}{{2a}}\ln \left| {a + x} \right| + \frac{1}{{2a}}\ln \left| {a - x} \right| + C \cr
& {\text{Using logarithmic properties}} \cr
& = \frac{1}{{2a}}\ln \left| {\frac{{a - x}}{{a + x}}} \right| + C \cr} $$
