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$$\int{\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}}}\mathrm dx$$

This was a Calc 2 problem for extra credit (we have done hyperbolic trig functions too, if that helps) and I didn't get it (don't think anyone did) -- how would you go about it?

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It's an elliptic integral, which I don't believe would have been taught in your course... –  Guess who it is. Dec 28 '10 at 6:50
So basically you'd say there's no way to integrate it using calc 2 knowledge? That's not a good extra credit... –  LinAlgStudent Dec 28 '10 at 6:51
If there is, it's not terribly obvious. Even if you try out a partial fraction decomposition, the three terms are still elliptic integrals (due to the $\sqrt{1+x^4}$ factor). –  Guess who it is. Dec 28 '10 at 6:58
Have you seen other integrals in your course that cannot be simplified (in terms of elemetary functions)? If so, the point of giving this problem may have been to remind you to be on the lookout for them. –  cch Dec 28 '10 at 7:09
@J.M: It seems like it might not be elliptic after all. See my answer... –  Aryabhata Dec 28 '10 at 7:51

4 Answers 4

up vote 49 down vote accepted

It might not be elliptic after all... (unless I have made some mistake)

$$\int{\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}}}\mathrm dx$$

Let $\displaystyle u = x -\frac{1}{x}$.

Then $\displaystyle du = (1 + \frac{1}{x^2})dx$.

Now $$\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}} = -\frac{x^2(1 + 1/x^2)}{x(x-1/x)\sqrt{x^2(x^2 + 1/x^2)}} = -\frac{1 + 1/x^2}{(x-1/x)\sqrt{(x - 1/x)^2 + 2}}$$


$$\int{\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}}}\mathrm dx$$

$$= -\int \ \frac{\mathrm{du}}{u \sqrt{u^2 + 2}}$$

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Hah, it is indeed not elliptic! $\frac1{\sqrt{2}}\ln\left(\frac{2x+\sqrt{2x^4+2}}{x^2-1}\right)$ does differentiate to the integrand! Great job; I'm glad I'm wrong! –  Guess who it is. Dec 28 '10 at 8:40
Both Maple and Mathematica are not smart enough to do this problem. –  TCL Dec 28 '10 at 15:32
Brilliant, thanks, so this is what we had to see. –  LinAlgStudent Dec 28 '10 at 16:09
@Moron: Ah, this the standard C.B.S.E 12th standard problem. –  user9413 May 7 '11 at 11:14
@TCL: Moron is smart enough, though. –  user9413 May 7 '11 at 11:15

Somewhat inspired by Moron's wonderful answer, I decided to see if a trigonometric solution would do the job.

Making the substitution $x=\cot\left(\frac{\theta}{2}\right)$, $\mathrm dx=\frac{\mathrm d\theta}{\cos\;\theta-1}$, we have

$$\int \frac{\mathrm d\theta}{\cos\;\theta(1-\cos\;\theta)\sqrt{1+\cot^4\frac{\theta}{2}}}$$

$$=\int \frac{\mathrm d\theta}{\cos\;\theta(1-\cos\;\theta)\sqrt{1+\left(\frac{1+\cos\;\theta}{1-\cos\;\theta}\right)^2}}$$

$$=\frac1{\sqrt{2}}\int \frac{\mathrm d\theta}{\cos\;\theta\sqrt{1+\cos^2\theta}}$$

which integrates to


Undoing the substitution, we get


and it is easy to verify that the derivative of this last expression gives the original integrand.

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Moron's and J.M.'s solutions are nice. Hopefully this solution is simpler.

Without loss of generality we may assume that $1\gt x\gt 0$. Put $x:=\sqrt{y}$, $1\gt y\gt 0$. Then we obtain $$ \int{\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}}}\mathrm dx=\int{\frac{1+y}{2(1-y)\sqrt{1+y^2}\sqrt{y}}}\mathrm dy. $$ Introduce the new variable $$ t:=\frac{1+y}{1-y},\qquad 1\lt t \lt \infty. $$ Then we have $$ y=\frac{-1+t}{1+t}, $$ $$ \mathrm dy=\frac{2}{(1+t)^2}\,\mathrm dt. $$ Substituting back we obtain $$ \int{\frac{1+y}{2(1-y)\sqrt{1+y^2}\sqrt{y}}}\mathrm dy=\int{\frac{t}{2\sqrt{1+\left(\dfrac{-1+t}{1+t} \right)^2}\sqrt{\dfrac{-1+t}{1+t}}}\frac{2}{(1+t)^2}\,\mathrm dt} $$ $$ =\frac{1}{\sqrt{2}}\int{\frac{t}{\sqrt{t^4-1}}}\mathrm dt $$ $$ =\frac{1}{2\sqrt{2}}\ln(t^2+\sqrt{t^4-1})+C. $$ Putting back everything we obtain $$ \frac{1}{2\sqrt{2}}\ln\left(\frac{(1+x^2)^2+2\sqrt{2}x\sqrt{1+x^4}}{(1-x^2)^2}\right)+C. $$

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What a surprise! Surfing the net, I found an almost same question on "hard integral" $$ \displaystyle \int \frac{x^2 - 1}{(x^2 + 1) \sqrt{x^4 + 1}} \, dx $$ from June 19, 2008.

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That seems to be very similar to my solution! I am not surprised though, after all, this appeared in a test, and I expect it to be well known in some circles. btw, if you liked this problem (which I am guessing you did :-)), you might like this too: math.stackexchange.com/questions/13414/… –  Aryabhata Jan 7 '11 at 18:00
Yes, I like to calculate integrals, although at the first step I check them by computer:-) . I also gave an elementary solution to your referred integral. –  vesszabo Jan 10 '11 at 14:28

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