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