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Evaluate $\int\frac1x(\frac{1-√x}{1+√x})^{\frac12}dx$

My Attempt:

Inside the bracket, multiplying and dividing by $(1-√x)$,

$\int\frac1x\frac{1-√x}{(1-x)^{\frac12}}dx$

Putting $(1-x)^\frac12=t\implies 1-x=t^2\implies x=1-t^2\implies dx=-2tdt$

Thus, the integral becomes $\int{\frac{1-\sqrt{1-t^2}}{(1-t^2)(t)}(-2t)dt}=-2\int{\frac{1-\sqrt{1-t^2}}{1-t^2}dt}$

It can be written as $-2\int{\frac1{1-t^2}dt}+2\int{\frac1{\sqrt{1-t^2}}dt}$

Thus, my answer is $-\ln|\frac{1+t}{1-t}|+2\arcsin t+c$, where $t=\sqrt{1-x}$

But the answer neither matches with wolfram nor with this video solution.

What's going on here?

EDIT:

After Tony's comment, I plotted my answer on wolfram, and it doesn't match with wolfram's answer to original question.

Wolfram also gives derivative of my answer, which sort of matches with the original question but not exactly.

I think the issue here is of domain.

But in general, while substituting, we don't think of domain. Or do we? If yes, is there any general advice which could be applied to such questions?

aarbee
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    I think you made a typo after your substitution as you should have $\int{\frac{1-\sqrt{1-t^2}}{(1-t^2)(t)}(-2t)dt}$ rather than $\int{\frac{2-\sqrt{1-t^2}}{(1-t^2)(t)}(-2t)dt}$. Then your answer would be $-\ln|\frac{1+t}{1-t}|+2\arcsin t+c$. I can't say if the Wolfram answer can be simplified to your one (the $\arctan$ is concerning), and the video is not loading for me. – Tony Mathew Aug 11 '23 at 05:10
  • @TonyMathew, yes, thanks, I'll correct the typo. – aarbee Aug 11 '23 at 05:12
  • One way to check without going through the algebra is plot your equation (for $c=0$) and the other solutions. If your one is parallel (due to different integration constant), then you can be re-assured that yours is correct. – Tony Mathew Aug 11 '23 at 05:13
  • @TonyMathew, ok, thanks for the input, I'll give it a try. – aarbee Aug 11 '23 at 05:15
  • @TonyMathew please check my edit. – aarbee Aug 11 '23 at 05:28
  • I view $\int f$ as some antiderivative of $f$ on some subset of the domain of $f$. I don't think it's too much to worry about. – Accelerator Aug 11 '23 at 06:46
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    Make life simpler $$\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}=t \implies x=\frac{\left(t^2-1\right)^2}{\left(t^2+1\right)^2}\implies dx=\frac{8 t \left(t^2-1\right)}{\left(t^2+1\right)^3},dt$$ The integral becomes $$\int \frac{8 t^2}{\left(t^2-1\right) \left(t^2+1\right)},dt$$

    Partial fraction decomposition leads to three simple integrals which recombine as one logarithm and one arctangent.

     – Claude Leibovici Aug 13 '23 at 01:24

1 Answers1

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Your solution does agree with WolframAlpha's (see the first listed alternate form):

$$\begin{align*} & -\ln\left|\frac{1+t}{1-t}\right| + 2 \arcsin t + C \\ &= - \ln\left|\frac{1+\sqrt{1-x}}{1-\sqrt{1-x}}\right| + 2\arcsin \sqrt{1-x} + C \\ &= - 2\operatorname{artanh} \sqrt{1-x} + 2\left(\frac\pi2-\arcsin \sqrt x\right) + C \\ &= -2\operatorname{artanh}\sqrt{1-x} - 2\arcsin \sqrt x + C \end{align*}$$

Make use of the identity, $$\arcsin x+\arcsin y=\arcsin\left( x\sqrt{1-y^2} + y\sqrt{1-x^2}\right)$$ as well as the logarithmic form of $\operatorname{artanh}$, $$\operatorname{artanh}z=\frac12 \ln\frac{1+z}{1-z}$$

user170231
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