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 for 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 of 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}

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