I have been asked about the following integral:

$$\int{\sqrt[4]{1-8{{x}^{2}}+8{{x}^{4}}-4x\sqrt{{{x}^{2}}-1}+8{{x}^{3}}\sqrt{{{x}^{2}}-1}}dx}$$ I think this is a joke of bad taste. I have tried every elementary method of integration which i know, also i tried integrating using Maple but as i suspected, the integrad doesn't have an anti derivative. Any ideas?

  • $\begingroup$ You could always put it into WolframAlpha, see if you get anything remotely nice and if so reverse engineer a solution from there. But be warned: some times WA gives ugly answers when there exist nice ones. $\endgroup$
    – Arthur
    Aug 1, 2018 at 4:33
  • $\begingroup$ WA doesn't give anything nice back: wolframalpha.com/input/… $\endgroup$
    – Mason
    Aug 1, 2018 at 4:36
  • 3
    $\begingroup$ All of these kinds of integrals are jokes in bad taste. The universe is laughing at the limited expressive strength of our mathematics. $\endgroup$
    – Daniel
    Aug 1, 2018 at 6:32
  • 3
    $\begingroup$ Can't the title be more descriptive? ......... $\endgroup$
    – user202729
    Aug 1, 2018 at 7:21
  • $\begingroup$ @Mason: WA often gives Wrong Answers, so even if it says something you still have to know how to verify it. $\endgroup$
    – user21820
    Aug 1, 2018 at 7:40

4 Answers 4


Use $$\sqrt[4]{1-8x^2+8x^4-4x\sqrt{x^2-1}+8x^3\sqrt{x^{2}-1}}=\left|x+\sqrt{x^2-1}\right|.$$

  • 4
    $\begingroup$ How did you discover that? $\endgroup$
    – Arthur
    Aug 1, 2018 at 4:47
  • 9
    $\begingroup$ @Arthur I tried to get $(a+b)^4$ under the radical and it turns out. $\endgroup$ Aug 1, 2018 at 4:50
  • 3
    $\begingroup$ Desmos confirms that the two functions are the same. $\endgroup$
    – Toby Mak
    Aug 1, 2018 at 4:57
  • $\begingroup$ It is a single hyperbola like curve with two disconnected parts. Each part can be obtained by changing sign in front of $\sqrt{...}$ $\endgroup$
    – Narasimham
    Aug 1, 2018 at 5:13
  • $\begingroup$ What made you think of that? Pls I need to know. $\endgroup$
    – William
    Aug 2, 2018 at 10:24


$(x+\sqrt{x^2-1})^4 = x^4+4x^3\sqrt{x^2-1}+6x^2(x^2-1)+4x(x^2-1)\sqrt{x^2-1}+(x^2-1)^2 = 8x^4-8x^2+1-4x\sqrt{x^2-1}+8x^3\sqrt{x^2-1}$



Let $\text{arcsec}x=t, x=\sec t$

Using Principal values, $0\le t\le\pi,t\ne\dfrac\pi2$

$\sin t=\sqrt{\left(1-\dfrac1x\right)^2}=\dfrac{\sqrt{x^2-1}}{|x|}$ as $\sin t\ge0$

$\tan t=\sin t\sec t=?$

$$1-8x^2+8x^4-4x\sqrt{x^2-1}+8x^3\sqrt{x^2-1}=\dfrac{\cos^4t-8\cos^2t+8-4\sin t\cos^2t+8\sin t}{\cos^4t}$$

Now $\cos^4t-8\cos^2t+8-4\sin t\cos^2t+8\sin t$

$=(1-\sin^2t)^2-8(1-\sin^2t)+8-4\sin t(1-\sin^2t)+8\sin t$

$=\sin^4t+4\sin^3t+6\sin^2t+4\sin t+1$

$=(\sin t+1)^4$


$$\frac{1}{16}{{\left( \sqrt{x-1}+\sqrt{x+1} \right)}^{8}}=1-8{{x}^{2}}+8{{x}^{4}}-4x\sqrt{{{x}^{2}}-1}+8{{x}^{3}}\sqrt{{{x}^{2}}-1}$$ So the integral becomes

$$\begin{align} & =\int{\sqrt[4]{\frac{1}{16}{{\left( \sqrt{x-1}+\sqrt{x+1} \right)}^{8}}}dx} \\ & =\frac{1}{2}\int{{{\left( \sqrt{x-1}+\sqrt{x+1} \right)}^{2}}dx} \\ & =\int{\left( x+\sqrt{{{x}^{2}}-1} \right)dx} \\ \end{align}$$

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    $\begingroup$ While this is nice, I want to warn people that $$f(x):=\frac{1}{2}\left(\sqrt{x-1}+\sqrt{x+1}\right)^2$$ is not the same as $$g(x):=\big|x+\sqrt{x^2-1}\big|\,.$$ Over the reals, $f(x)$ is defined only for $x\geq 1$, but $g(x)$ is defined for all $x$ with $|x|\geq 1$. I subtracted points in the past from students who had written something like this in their homework or exam papers. (But don't worry, I am not downvoting you.) $\endgroup$ Aug 1, 2018 at 8:55

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