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So while working on some physics problem for differential equations, I landed at this weird integral $$ \int \frac 1 {\sqrt{1-\left(\frac 2x\right)}}\,dx $$

So since there is a square root, I thought I could use trig substitution, but I couldn't find anything that works out. How can one solve this integral in a nice simple manner? If you can solve it in a different way, it is still fine. $Thank$ $you!$ This is the answer given to me by symbolab $$ 4\left(-\frac{1}{4\left(\sqrt{1-\frac{2}{x}}-1\right)}-\frac{1}{4\left(\sqrt{1-\frac{2}{x}}+1\right)}-\frac{1}{4}\ln \left|\sqrt{1-\frac{2}{x}}-1\right|+\frac{1}{4}\ln \left|\sqrt{1-\frac{2}{x}}+1\right|\right)+C $$

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HINT:

Make the substitution $x=2\sec^2(\theta)$ and arrive at

$$4\int \sec^3(\theta)\,d\theta$$

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  • $\begingroup$ Why did this receive a down vote? Would the down voter care to state a reason? $\endgroup$
    – Mark Viola
    Feb 18 '16 at 21:09
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$u=\sqrt{1-\frac{2}{x}} \implies u^2=1-\frac{2}{x} \implies 2u du=\frac{2}{x^2} dx \implies du=\frac{1}{x^2} \frac{1}{u} dx \implies du=\frac{1}{x^2} \frac{1}{\sqrt{1-\frac{2}{x}}} dx \text{ So now back to the integral } \int \frac{1}{\sqrt{1-\frac{2}{x}}}dx=\int \frac{x^2}{x^2 \sqrt{1-\frac{2}{x}}}dx \\ \text{ now putting in the sub } \int (\frac{2}{1-u^2})^2 du \text{ Note: I solve for }x^2 \text{ from my sub to plug in also } \\ \text{ That is I solved } u^2=1-\frac{2}{x} \text{ for } x \text{ then squared result } \\ \text{ Then you can do partial fractions. } $

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Adding to Dr. MV's hint, $$\int\frac{1}{\sqrt{1-\frac2x}}\ \mathrm dx=\int\frac{\sqrt x}{\sqrt{x-2}}\ \mathrm dx$$ Using trigonometric substitution, we have $$\sqrt x=\sqrt 2\sec\phi$$ Which implies that $$x=2\sec^2\phi\Rightarrow \mathrm dx=4\tan\phi\sec^2\phi\ \mathrm d\phi$$ So now the integral becomes $$\int\frac{4\sqrt 2\tan\phi\sec^3\phi}{\sqrt{2\sec^2\phi-2}}\ \mathrm d\phi$$ $$=4\int\frac{\tan\phi\sec^3\phi}{\sqrt{\sec^2\phi-1}}\ \mathrm d\phi$$ $$=4\int\frac{\tan\phi\sec^3\phi}{\sqrt{\tan^2\phi}}\ \mathrm d\phi$$ $$=4\int\sec^3\phi\ \mathrm d\phi$$ This integral has been answered numerous times. Study this to see how the following solution is derived. $$2\sec\phi\tan\phi+2\ln\left|\sec\phi+\tan\phi\right|+C$$ Now let's get an answer in terms of $x$. Note that $$\sqrt x=\sqrt 2\sec\phi\Rightarrow\phi=\mathrm{arcsec}\left(\frac{\sqrt x}{\sqrt 2}\right)$$ And $$ \tan\left(\mathrm{arcsec}(x)\right)=\sqrt{x^2-1}$$ Therefore $$\frac{2\sqrt x}{\sqrt 2}\tan\left(\mathrm{arcsec}\left(\frac{\sqrt x}{\sqrt 2}\right)\right)+2\ln\left|\frac{\sqrt x}{\sqrt 2}+\tan\left(\mathrm{arcsec}\left(\frac{\sqrt x}{\sqrt 2}\right)\right)\right|+C$$ $$=\sqrt{x^2-2x}+2\ln\left|\sqrt{x-2}+\sqrt x\right|+C$$

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  • $\begingroup$ No its fine I know the rest. Thank you! $\endgroup$
    – Michael
    Feb 16 '16 at 12:21
  • $\begingroup$ @Michael, glad I could help. $\endgroup$
    – k170
    Feb 16 '16 at 14:26
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Some hints

First of all, simple, thing, just rewrite the integrand this way

$$\frac{1}{\sqrt{\frac{x-2}{x}}}$$

Then substitute

$$y = \frac{x-2}{x} ~~~~~~~ \text{d}y = \frac{1}{x} - \frac{x-2}{x^2}\ \text{d}x$$

You're now with

$$2\int \frac{1}{(1 + y^2)\sqrt{y}}\ \text{d}y$$

And for this one, use

$$z = \sqrt{y} ~~~~~~~ \text{d}z = \frac{1}{2\sqrt{y}}\ \text{d}y$$

and your integral is now:

$$4\int \frac{1}{(1 - z^2)^2}\ \text{d}z$$

which is easily solvable by partial fractions:

$$\frac{1}{(1 - s^2)^2} = \frac{1}{4(z+1)} + \frac{1}{4(z+1)^2} - \frac{1}{4(z-1)} + \frac{1}{4(z-1)^2}$$

Split and do the easy integrals.

Final Result

$$x\sqrt{\frac{x-2}{x}} - \ln\left(\sqrt{\frac{x-2}{x}} - 1\right) + \ln\left(\sqrt{\frac{x-2}{x}} + 1\right) + C$$

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    $\begingroup$ It seems to me that rewriting $1-\frac2x$ as $\frac{x-2}x$ complicates the problem rather than simplifying it. Among other things, isn't it easier to write $\text dy = \frac2{x^2}\ \text dx$ instead of $\text dy = \frac1x - \frac{x-2}x\ \text dx$? $\endgroup$
    – Théophile
    Feb 16 '16 at 20:03

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