Integral of $\int \frac{\sqrt{x^2-1}}{x}dx$
Solution 1:
$$\int \:\frac{\sqrt{x^2-1}}{x}dx$$
Let $u=\:\sqrt{x^2-1}$ then
$$\frac{du}{dx}=\frac{x}{\sqrt{x^2-1}}\iff dx=\frac{\sqrt{x^2-1}}{x}\:du$$
Thus
\begin{align*}
\int \:\frac{\sqrt{x^2-1}}{x}dx& =\int \:\frac{u^2}{u^2+1}du\\
&=\int \:\frac{u^2 + 1 - 1}{u^2+1 + 1 - 1}du\\
&=\int \:\left(\frac{u^2+1}{u^2+1}-\frac{1}{u^2+1}\right)du\\
&=\int \:\left(1-\frac{1}{u^2+1}\right)du\\
&=u-\arctan \left(u\right)+C\\
&=\:\sqrt{x^2-1}-\arctan \left(\:\sqrt{x^2-1}\right)+C
\end{align*}
Solution 2:
The answer is $$\int \frac{\sqrt{x^{2}-1}}{x}{\rm d}x= \sqrt{x^{2}-1}-\arctan\left( \sqrt{x^{2}-1}\right)+C.$$ In the integral perform the change of variable $x=\sec u$, \begin{align*} \int\frac{\sqrt{x^{2}-1}}{x}{\rm d}x&= \int \frac{\sqrt{\sec^{2}u-1}}{\sec u}\tan u\sec u{\rm d}u\\ &=\int \tan u \tan u{\rm d}u\\ &=\int \tan^{2} u{\rm d}u\\ &=\int \sec^{2}u -1 {\rm d}u\\ &=\tan(u)-u+C \end{align*} Therefore, \begin{align*} \int \frac{\sqrt{x^{2}-1}}{x}{\rm d}x&=\tan(u)-u+C\\ &= \tan\left( \sec^{-1}(x)\right)-\sec^{-1}(x)+C\\ &=\sqrt{x^{2}-1}-\sec^{-1}(x)+C\\ &=\boxed{\sqrt{x^{2}-1}-\arctan\left( \sqrt{x^{2}-1}\right)+C} \end{align*}