Computing the integral $\int \sqrt{\frac{x-1}{x+1}}\,\mathrm dx $

Let $\sqrt{\dfrac{x-1}{x+1}} = t$. We then get that $$\dfrac{x-1}{x+1} = t^2 \implies x-1 = t^2(x+1) \implies x = \dfrac{1+t^2}{1-t^2} \implies dx = \dfrac{4t}{(1-t^2)^2}dt$$ Hence, we get that $$\int \sqrt{\dfrac{x-1}{x+1}} dx = \int \dfrac{4t^2}{(1-t^2)^2} dt$$ I trust you can take it from here via the method of partial fractions.