Integration of 1 Over a^2-x^2

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

The integration is of the form

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

Now we have an integral to evaluate,

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