# How do I compute the next integral: int sqrt( (x-1)/(x+1))dx ?

How do I compute the next integral:
$\int \sqrt{\frac{x-1}{x+1}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x\phantom{\rule{thickmathspace}{0ex}}?$
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Warkallent8
Let $\sqrt{\frac{x-1}{x+1}}=t$. We then get that
$\frac{x-1}{x+1}={t}^{2}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}x-1={t}^{2}\left(x+1\right)\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}x=\frac{1+{t}^{2}}{1-{t}^{2}}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}dx=\frac{4t}{\left(1-{t}^{2}{\right)}^{2}}dt$
Hence, we get that
$\int \sqrt{\frac{x-1}{x+1}}dx=\int \frac{4{t}^{2}}{\left(1-{t}^{2}{\right)}^{2}}dt$
I trust you can take it from here via the method of partial fractions.