# Integration of 1 Over a^2+x^2

In this tutorial we shall discuss the integration of 1 over x^2+a^2, and this is another important form of integration.

The integration of $\frac{1}{{{a^2} + {x^2}}}$ is of the form
$\int {\frac{1}{{{a^2} + {x^2}}}dx = } \frac{1}{a}{\tan ^{ – 1}}\left( {\frac{x}{a}} \right) + c$

To prove this formula, putting $x = a\tan t$ we have $dx = a{\sec ^2}tdt$, $t = {\tan ^{ – 1}}\left( {\frac{x}{a}} \right)$. So the given integral takes the form
$\begin{gathered} \int {\frac{{dx}}{{{a^2} + {x^2}}} = \int {\frac{{a{{\sec }^2}tdt}}{{{a^2} + {a^2}{{\tan }^2}t}}} } \\ \Rightarrow \int {\frac{{dx}}{{{a^2} + {x^2}}} = \int {\frac{{a{{\sec }^2}tdt}}{{{a^2}\left( {1 + {{\tan }^2}t} \right)}}} } \\ \Rightarrow \int {\frac{{dx}}{{{a^2} + {x^2}}} = \int {\frac{{a{{\sec }^2}t}}{{{a^2}{{\sec }^2}t}}dt} } \\ \Rightarrow \int {\frac{{dx}}{{{a^2} + {x^2}}} = \frac{1}{a}\int {dt} } \\ \Rightarrow \int {\frac{{dx}}{{{a^2} + {x^2}}} = \frac{1}{a}t + c} \\ \end{gathered}$

Using the value $t = {\tan ^{ – 1}}\left( {\frac{x}{a}} \right)$, we have
$\Rightarrow \int {\frac{{dx}}{{{a^2} + {x^2}}} = \frac{1}{a}{{\tan }^{ – 1}}\left( {\frac{x}{a}} \right) + c}$