Integration of 1 Over x^2-a^2

In this tutorial we shall find the integral of 1 over x^2-a^2.

The integration is of the form

\int {\frac{1}{{{x^2} - {a^2}}}dx} = \frac{1}{{2a}}\ln \left( {\frac{{x - a}}{{x + a}}} \right) + c

Now we have an integral to evaluate,

\begin{gathered} I = \int {\frac{1}{{{x^2} - {a^2}}}dx} \\ \Rightarrow I = \int {\frac{1}{{\left( {x - a} \right)\left( {x + a} \right)}}dx} \\ \Rightarrow I = \frac{1}{{2a}}\int {\frac{{\left[ {\left( {x + a} \right) - \left( {x - a} \right)} \right]}}{{\left( {x - a} \right)\left( {x + a} \right)}}dx} \\ \Rightarrow \int {\frac{{dx}}{{{x^2} - {a^2}}} = \frac{1}{{2a}}\left[ {\int {\frac{1}{{x - a}}dx - \int {\frac{1}{{x + a}}dx} } } \right]} \\ \end{gathered}

Using the integral formula \int {\frac{{f'\left( x \right)}}{{f\left( x \right)}}dx = \ln f\left( x \right) + c} , we have

\begin{gathered} \int {\frac{{dx}}{{{x^2} - {a^2}}} = \frac{1}{{2a}}\left[ {\ln \left( {x - a} \right) - \ln \left( {x + a} \right)} \right]} + c \\ \Rightarrow \int {\frac{{dx}}{{{x^2} - {a^2}}} = \frac{1}{{2a}}\ln \frac{{x - a}}{{x + a}} + c} \\ \end{gathered}