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

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

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

To prove this formula, consider
$\frac{d}{{dx}}\left[ {{{\sec }^{ – 1}}x + c} \right] = \frac{d}{{dx}}{\sec ^{ – 1}}x + \frac{d}{{dx}}c$

Using the derivative formula $\frac{d}{{dx}}{\sec ^{ – 1}}x = \frac{1}{{x\sqrt {{x^2} – 1} }}$, we have
$\begin{gathered} \frac{d}{{dx}}\left[ {{{\sec }^{ – 1}}x + c} \right] = \frac{1}{{x\sqrt {{x^2} – 1} }} + 0 \\ \Rightarrow \frac{d}{{dx}}\left[ {{{\sec }^{ – 1}}x + c} \right] = \frac{1}{{x\sqrt {{x^2} – 1} }} \\ \Rightarrow \frac{1}{{x\sqrt {{x^2} – 1} }} = \frac{d}{{dx}}\left[ {{{\sec }^{ – 1}}x + c} \right] \\ \Rightarrow \frac{1}{{x\sqrt {{x^2} – 1} }}dx = d\left[ {{{\sec }^{ – 1}}x + c} \right]\,\,\,\,{\text{ – – – }}\left( {\text{i}} \right) \\ \end{gathered}$

Integrating both sides of equation (i) with respect to $x$, we have
$\int {\frac{1}{{x\sqrt {{x^2} – 1} }}dx} = \int {d\left[ {{{\sec }^{ – 1}}x + c} \right]}$

As we know that by definition, integration is the inverse process of the derivative, so we have
$\int {\frac{1}{{x\sqrt {{x^2} – 1} }}dx} = {\sec ^{ – 1}}x + c$

We know that the derivative of $\frac{d}{{dx}}{\csc ^{ – 1}}x = – \frac{1}{{x\sqrt {{x^2} – 1} }}$

This formula can also be written as
$\int {\frac{1}{{x\sqrt {{x^2} – 1} }}dx} = – {\csc ^{ – 1}}x + c$