$$\int \sqrt{\frac{x-1}{x+1}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x\phantom{\rule{thickmathspace}{0ex}}?$$

Aarav Atkins
2022-10-22
Answered

How do I compute the next integral:

$$\int \sqrt{\frac{x-1}{x+1}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x\phantom{\rule{thickmathspace}{0ex}}?$$

$$\int \sqrt{\frac{x-1}{x+1}}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x\phantom{\rule{thickmathspace}{0ex}}?$$

You can still ask an expert for help

Warkallent8

Answered 2022-10-23
Author has **16** answers

Let $\sqrt{{\displaystyle \frac{x-1}{x+1}}}=t$. We then get that

$$\frac{x-1}{x+1}}={t}^{2}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}x-1={t}^{2}(x+1)\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}x={\displaystyle \frac{1+{t}^{2}}{1-{t}^{2}}}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}dx={\displaystyle \frac{4t}{(1-{t}^{2}{)}^{2}}}dt$$

Hence, we get that

$$\int \sqrt{{\displaystyle \frac{x-1}{x+1}}}dx=\int {\displaystyle \frac{4{t}^{2}}{(1-{t}^{2}{)}^{2}}}dt$$

I trust you can take it from here via the method of partial fractions.

$$\frac{x-1}{x+1}}={t}^{2}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}x-1={t}^{2}(x+1)\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}x={\displaystyle \frac{1+{t}^{2}}{1-{t}^{2}}}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}dx={\displaystyle \frac{4t}{(1-{t}^{2}{)}^{2}}}dt$$

Hence, we get that

$$\int \sqrt{{\displaystyle \frac{x-1}{x+1}}}dx=\int {\displaystyle \frac{4{t}^{2}}{(1-{t}^{2}{)}^{2}}}dt$$

I trust you can take it from here via the method of partial fractions.

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