Question

Integrate $\int \frac{\sqrt{{\mathrm{x}}^2-1}}{\mathrm{x}}\mathrm{d}x$.

Answer 1

Step 1: Use Integration by substitution

Given, $\int \frac{\sqrt{{\mathrm{x}}^2-1}}{\mathrm{x}}\mathrm{d}x$

Let, $I=\int \frac{\sqrt{{\mathrm{x}}^2-1}}{\mathrm{x}}\mathrm{d}x$

Let, $x=\sec \left(t\right)$
Differentiating on both sides,

$\begin{array}{cc}& \frac{\mathrm{d}x}{\mathrm{d}t}=\sec \mathit{\left(}\mathit{t}\mathit{\right)}\tan \left(t\right)\\ \Rightarrow & \mathrm{d}x=\sec \mathit{\left(}t\mathit{\right)}\tan \left(t\right)\mathrm{d}t\end{array}$

Step 2: Substituting the values in $I$,

$\begin{array}{ccc}\Rightarrow & I=\int \frac{\sqrt{{\sec }^{\mathit{2}}\left(t\right)-1}}{\sec \left(t\right)}\sec \left(t\right)\tan \left(t\right)\mathrm{d}t& \\ \Rightarrow & I=\int {\tan }^2\left(t\right)dt& \left[\because {\sec }^2\left(t\right)={\tan }^2\left(t\right)+1\right]\\ \Rightarrow & I=\int \left[{\sec }^2\left(\mathrm{t}\right)-1\right]\mathrm{d}t& \\ \Rightarrow & I=\int {\sec }^2\left(t\right)\mathrm{d}t-\int 1\mathrm{d}t& \\ \Rightarrow & I=\tan \left(t\right)-t+C& \left[\because \int se{c}^2\left(t\right)dt=tan\left(t\right)\right]\end{array}$

Back substituting,

$\begin{array}{clc}\Rightarrow & I=\tan \left({\sec }^{-1}\left(x\right)\right)-{\sec }^{-1}\left(x\right)+C& \\ \Rightarrow & I=\tan \left({\tan }^{-1}\left(\sqrt{x^2-1}\right)\right)-se{c}^{-1}\left(x\right)+C& \left[\because \tan \left({\tan }^{-1}\left(x\right)\right)=x\right]\\ \Rightarrow & I=\sqrt{x^2-1}-se{c}^{-1}\left(x\right)+C& \end{array}$

Therefore, $\int \frac{\sqrt{x^2-1}}{x}\mathrm{d}x=\sqrt{x^2-1}-{\sec }^{-1}\left(x\right)+\mathrm{C}$.