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$$\int \dfrac {\operatorname d\!x} {2x \sqrt{1-x}\sqrt{2-x + \sqrt{1-x}}}$$

Hey there, I've got this complicated integral to evaluate, but I don't know how to go about. I have tried making two substitutions:

  1. $ t^2 = 1 - x $

  2. $ x = \sin^2\theta $

But both gave another complicated integral to evaluate:

$$ \int \dfrac {\operatorname d\!t} {(t^2-1)\sqrt{ t^2 + t + 1 }} $$

I tried to get the answer for this one using wolfram alpha, but it gave a HUGE, simply HUGE solution. I also tried to get the solution for the original question via wolfram alpha, but it timed out.

Any ideas?

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  • $\begingroup$ Where does the integral come from? Do you need the full analytic solution? $\endgroup$ May 8, 2013 at 14:43
  • $\begingroup$ We had this as a test question. But now I have started doubting that there might be some error in the question, don't know though because no one had announced that there was any error in the question paper. $\endgroup$ May 8, 2013 at 14:44
  • $\begingroup$ No, hints will do, as they almost always do in my questions :D (don't know what you mean by full analytic solution though!) $\endgroup$ May 8, 2013 at 14:46
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    $\begingroup$ Mathematica gives for first integral this: $\frac12\left(\frac{\log(1-\sqrt{1-x})}{\sqrt3}-\log(1+\sqrt{1-x})+\log(1-\sqrt{1-x}+2\sqrt{2+\sqrt{1-x}-x})-\frac{\log(3+3\sqrt{1-x}+2\sqrt3\sqrt{2+\sqrt{1-x}-x})}{\sqrt3}\right)$ $\endgroup$
    – Ruslan
    May 8, 2013 at 14:49
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    $\begingroup$ You are getting a log. Means : USE partial fractions. Seems absurd, though. $\endgroup$
    – Inceptio
    May 8, 2013 at 14:53

3 Answers 3

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For what it's worth, you can break things up a bit by recognizing that

$$\frac{1}{t^2-1} = \frac12 \left ( \frac{1}{t-1} - \frac{1}{t+1}\right )$$

For example, consider the $t-1$ piece; you may substitute $u=t+1/2$ and get

$$\int \frac{dt}{(t-1) \sqrt{t^2+t+1}} = \int \frac{du}{(u-3/2) \sqrt{u^2+3/4}}$$

This latter integral is relatively tame according to WA:

$$ \frac{1}{\sqrt{3}} \left[\log{\left(u-\frac{3}{2}\right)}-\log{\left(\sqrt{12 u^2+9}+3 u+\frac{3}{2}\right)}\right]+C$$

where $C$ is a constant of integration. A similar expression may be found for the $t+1$ piece.

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  • $\begingroup$ I think that's a good way to go...! $\endgroup$ May 8, 2013 at 15:41
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The following integral

$$ \int \frac{1}{x \sqrt{ax^2 + bx + c}}\text{d}x$$

can be solved by substituting $x = \frac{1}{t}$ to get

$$\int \frac{\pm 1}{\sqrt{a + bt + ct^2}}\text{d}t$$

Which can be recast as the derivative of an inverse trigonometric function (could be hyperbolic, depending on the signs taken):

$$\int \frac{\pm 1}{\sqrt{\pm 1 \pm x^2}}$$

And as noted by Ron, your integral is the sum of two such integrals.

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The $t$-integral is primed for the well-known Euler substitution. Let

$$u=\frac{\sqrt{t^2+t+1} - 1}t \implies t = -\frac{1-2u}{1-u^2} \implies dt = \frac{2u^2-2u+2}{(1-u^2)^2} \, du$$

to transform to

$$\begin{align*} I &= \int \dfrac{dt}{(t^2-1)\sqrt{ t^2 + t + 1 }} \\[1ex] &= 2 \int \frac{u^2 - u + 1}{(1-u^2)^2 \left(\frac{(1-2u)^2}{(1-u^2)^2} - 1\right) \left(1-\frac{u-2u^2}{1-u^2}\right)} \, du \\[1ex] &= 2 \int \frac{u^2-1}{u(u-2)(u^2+2u-2)} \, du \end{align*}$$

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