# Integration of 1 Over (x Square Root of (x^2-a^2))

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

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

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

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