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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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    $\begingroup$ It's an elliptic integral, which I don't believe would have been taught in your course... $\endgroup$ Dec 28, 2010 at 6:50
  • $\begingroup$ So basically you'd say there's no way to integrate it using calc 2 knowledge? That's not a good extra credit... $\endgroup$ Dec 28, 2010 at 6:51
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    $\begingroup$ 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). $\endgroup$ Dec 28, 2010 at 6:58
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    $\begingroup$ 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. $\endgroup$
    – cch
    Dec 28, 2010 at 7:09
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    $\begingroup$ @J.M: It seems like it might not be elliptic after all. See my answer... $\endgroup$
    – Aryabhata
    Dec 28, 2010 at 7:51

6 Answers 6

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

Thus

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

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

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    $\begingroup$ 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! $\endgroup$ Dec 28, 2010 at 8:40
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    $\begingroup$ Both Maple and Mathematica are not smart enough to do this problem. $\endgroup$
    – TCL
    Dec 28, 2010 at 15:32
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    $\begingroup$ Brilliant, thanks, so this is what we had to see. $\endgroup$ Dec 28, 2010 at 16:09
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    $\begingroup$ @Moron: Ah, this the standard C.B.S.E 12th standard problem. $\endgroup$
    – user9413
    May 7, 2011 at 11:14
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    $\begingroup$ @TCL: Moron is smart enough, though. $\endgroup$
    – user9413
    May 7, 2011 at 11:15
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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

$$\frac1{\sqrt{2}}\tanh^{-1}\frac{\sin\;\theta}{\sqrt{1+\cos^2\theta}}$$

Undoing the substitution, we get

$$\frac1{\sqrt{2}}\tanh^{-1}\left(x\sqrt{\frac{2}{x^4+1}}\right)$$

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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    $\begingroup$ 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/… $\endgroup$
    – Aryabhata
    Jan 7, 2011 at 18:00
  • $\begingroup$ 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. $\endgroup$
    – vesszabo
    Jan 10, 2011 at 14:28
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\begin{align}\int \frac{(1+x^2)dx}{(1-x^2)\sqrt{x^4+1}} =&\int \frac{\frac{(1+x^2)dx}{(1-x^2)|1-x^2|}}{\sqrt{\frac{x^4+1}{(1-x^2)^2}}} =\int \frac{d(\frac x{|1-x^2|})}{\sqrt{1+\frac{2x^2}{(1-x^2)^2}}} =\frac1{\sqrt2}\sinh^{-1}\frac{\sqrt2 x}{|1-x^2|} \end{align}

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$$\begin{align*} & \int\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}}\,dx \\ &= \int\frac{y}{\sqrt{1+\frac{(y-1)^2}{(y+1)^2}}}\cdot\frac{dy}{(y+1)\sqrt{y^2-1}} \tag1\\ &= \frac1{\sqrt2} \int\frac{y}{\sqrt{y^4-1}}\,dy \\ &= \frac1{2\sqrt2} \int\frac{dz}{\sqrt{z^2-1}} \tag2 \\ &= \frac1{2\sqrt2} \tanh^{-1}\left(\frac z{\sqrt{z^2-1}}\right) + C \\ &= \frac1{2\sqrt2} \tanh^{-1}\left(\frac{y^2}{\sqrt{y^4-1}}\right) + C \\ &= \frac1{2\sqrt2} \tanh^{-1}\left(\frac{\left(\frac{1+x^2}{1-x^2}\right)^2}{\sqrt{\left(\frac{1+x^2}{1-x^2}\right)^4-1}}\right) + C \\ &= \frac1{2\sqrt2} \tanh^{-1}\left(\frac{(1+x^2)^2}{\sqrt{8x^2(1+x^4)}}\right) + C \end{align*}$$


  • $(1)$ : substitute $y = \dfrac{1+x^2}{1-x^2} \color{#0000006F}{\iff x=\sqrt{\dfrac{y-1}{y+1}} \implies dx = \dfrac{dy}{(y+1)\sqrt{y^2-1}}}$
  • $(2)$ : substitute $z=y^2 \color{#0000006F}{\iff y=\sqrt z \implies dy=\dfrac{dz}{2\sqrt z}}$
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