Answer
$$\frac{1}{2}\left[ {\ln \left| {1 + x} \right| - \ln \left| {1 - x} \right|} \right] + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{1 - {x^2}}}} \cr
& {\text{The form of the partial fraction decomposition is}} \cr
& \frac{1}{{1 - {x^2}}} = \frac{1}{{\left( {1 - x} \right)\left( {1 + x} \right)}} = \frac{A}{{1 - x}} + \frac{B}{{1 + x}} \cr
& \cr
& {\text{Multiplying by }}\left( {1 - x} \right)\left( {1 + x} \right){\text{, we have}} \cr
& {\text{1}} = A\left( {1 + x} \right) + B\left( {1 - x} \right) \cr
& {\text{if we set }}x = 1, \cr
& {\text{1}} = A\left( {1 + 1} \right) \cr
& A = \frac{1}{2} \cr
& {\text{if we set }}x = - 1, \cr
& {\text{1}} = B\left( 2 \right) \cr
& B = \frac{1}{2} \cr
& {\text{then}} \cr
& \frac{1}{{1 - {x^2}}} = \frac{{1/2}}{{1 - x}} + \frac{{1/2}}{{1 + x}} \cr
& \int {\frac{{dx}}{{1 - {x^2}}}} = \int {\left( {\frac{1}{{2\left( {1 - x} \right)}} + \frac{1}{{2\left( {1 + x} \right)}}} \right)dx} \cr
& {\text{integrating}} \cr
& = - \frac{1}{2}\ln \left| {1 - x} \right| + \frac{1}{2}\ln \left| {1 + x} \right| + C \cr
& = \frac{1}{2}\left[ {\ln \left| {1 + x} \right| - \ln \left| {1 - x} \right|} \right] + C \cr} $$
