evaluate integral the simplest way possible I want to evaluate the integration $\int \limits x \sqrt{\frac{x-1}{x+1}}dx$ ?

I tried putting $t = \sqrt{\frac{x-1}{x+1}}$ but that did not help !

Is there a quick and smart way to evaluate.
For the square root to be defined you need for instance $x\ge 1$, this suggest a substitution $x=\cosh(u)$.

The integral becomes $$I=\int\cosh(u)\sinh(u)\sqrt{\frac{\cosh(u)-1}{\cosh(u)+1}}\mathop{du}$$

Now using $\cosh(u)^2-\sinh(u)^2=1$ we get 



$\displaystyle\int\frac{\cosh(u)\sinh(u)\sqrt{\cosh(u)^2-1}}{\cosh(u)+1}\mathop{du}=\int\frac{\cosh(u)\sinh(u)^2}{\cosh(u)+1}\mathop{du}=\int\cosh(u)^2-\cosh(u)\mathop{du}$



Just linearise the $\cosh(u)^2$ to get the result $$I=\frac u2-\sinh(u)+\frac 14\sinh(2u)$$


Finally use $\sinh(\cosh^{-1}(x))=\sqrt{x^2-1}\quad$ and $\quad\sinh(2\cosh^{-1}(x))=2x\sqrt{x^2-1}$

To get $$I=\frac 12\cosh^{-1}(x)+\sqrt{x^2-1}\left(\frac x2-1\right)+C$$
Inside the radical, multiply top and bottom by $x-1$ to get

$$\int \frac{x^2-x}{\sqrt{x^2-1}} \; dx$$

This wants the usual trig sub $x = \sec \theta$ which yields

$$\int \sec^3\theta - \sec^2 \theta \; d\theta.$$

The $\sec^2 \theta$ is easy.  The $\sec^3\theta$ is less easy, but is a standard integral.