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

In this tutorial we shall discuss the integration of 1 over x into square root of x^2-1, and this is another important for 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 derivative, so we have

\int  {\frac{1}{{x\sqrt {{x^2} - 1} }}dx}  =  {\sec ^{ - 1}}x + c


As we know that 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

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