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A recent post in Mathematics Stack Exchange claims that one can get from Mathematica the following result: $$\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{dx dy dz}{(1+x^2+y^2+z^2)^2}=\frac{\pi^2}{32}. $$

However, I am not able to reproduce the result with Mathematica 11. With the code

Integrate[1/(1+x^2+y^2+z^2)^2,{x,0,1},{y,0,1},{z,0,1}]

I only got:

enter image description here

With WolframAlpha, one has a numerical result:

enter image description here

Question: Can Mathematica calculate the mentioned triple integral?


[Added later:] According to what the author later added in his post, he actually calculated the integral with Mathematica only numerically. Lele's answer below thus answers the question in the title: yes, the triple integral can be calculated, numerically. I am also curious to know though if Mathematica can return the result as $\dfrac{\pi^2}{32}$.

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    $\begingroup$ Please post the code you used to attempt to do this integral. $\endgroup$ – bbgodfrey Sep 11 at 17:24
  • $\begingroup$ @bbgodfrey: it is nothing but one line and it is in the screen shot. Fair enough. I have now included it in the post. Thanks for your comment. $\endgroup$ – Jack Sep 11 at 17:41
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    $\begingroup$ Version 12 gives essentially the same answer, Integrate[ArcCot[Sqrt[2 + x^2]]/((1 + x^2)*Sqrt[2 + x^2]), {x, 0, 1}]. I have no idea why the question was down-voted. $\endgroup$ – bbgodfrey Sep 11 at 18:07
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For numerical integrations, you should use NIntegrate instead of Integrate.

With

NIntegrate[1/(1 + x^2 + y^2 + z^2)^2, {x, 0, 1}, {y, 0, 1}, {z, 0, 1}]

you obtain the same result as in WolframAlpha:

enter image description here

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  • $\begingroup$ A symbolic answer appears to be desired. $\endgroup$ – bbgodfrey Sep 11 at 17:25
  • $\begingroup$ Ok, the OP has changed the question now $\endgroup$ – Lele Sep 11 at 17:44
  • $\begingroup$ @Lele: +1 Thanks for your answer and sorry for the ambiguity. I will try to find better way to edit my post. $\endgroup$ – Jack Sep 11 at 17:48

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