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
\[\int {\frac{1}{{x\sqrt {{x^2} – 1} }}dx} = – {\csc ^{ – 1}}x + c\]