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I have built this integral with the purpose of presenting a question. I find interesting and pleasant to readers MSE for the New Year 2016 and expecting to see different methods of solution.

I can confirm that this has been the case by the welcome that has been given and the two motivated answers it have had.

Calculate:

$$\large \int_{2016}^{3\cdot 2016}\frac{\sqrt[5]{3\cdot 2016-x} }{\sqrt[5]{3\cdot 2016-x}+\sqrt[5]{x-2016}}\mathrm dx$$

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    $\begingroup$ After a standard change of variables, the answer seems to be 2016. $\endgroup$
    – Hmm.
    Dec 31, 2015 at 18:46
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    $\begingroup$ @SoumyaSinhaBabu: Go ahead dear friend. $\endgroup$
    – Piquito
    Dec 31, 2015 at 18:48
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    $\begingroup$ Observation - $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x)dx$. $\endgroup$
    – Hmm.
    Dec 31, 2015 at 18:51
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    $\begingroup$ Is this a challenge, where you know the answer? Please provide more context. $\endgroup$
    – Rory Daulton
    Dec 31, 2015 at 20:57
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    $\begingroup$ It was in fact a challenge, a pleasant and pertinent one I feel. Tomorrow I´ll try to edit with my deficient English (I built the question myself for the New Year). $\endgroup$
    – Piquito
    Jan 1, 2016 at 1:56

2 Answers 2

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Define $$I:=\int_{k}^{3k}\frac{\left(3k-x\right)^{1/5}dx}{\left(3k-x\right)^{1/5}+\left(x-k\right)^{1/5}},\,k:=2016.$$

The substitution $x\mapsto 4k-x$ gives$$I:=\int_{k}^{3k}\frac{\left(x-k\right)^{1/5}dx}{\left(3k-x\right)^{1/5}+\left(x-k\right)^{1/5}}.$$Halving the sum of these expressions gives$$I=\frac12\int_k^{3k}dx=k.$$

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    $\begingroup$ A more complete answer than the one posted above, as you have generailsed to a limit $k$. nice =1 $\endgroup$
    – user284001
    Oct 18, 2016 at 10:16
  • $\begingroup$ I see the notation := often; what does it mean? $\endgroup$
    – user71207
    Apr 3, 2021 at 11:59
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    $\begingroup$ @user71207 "is defined as". You can also write a definition backwards, viz. $\prod_{j=1}^nx=:x^n$, but that's a lot rarer. $\endgroup$
    – J.G.
    Apr 3, 2021 at 12:23
  • $\begingroup$ Ok thanks. Do you still have to say "Define $I := blah blah$" If := already means "define this to be"? $\endgroup$
    – user71207
    Apr 4, 2021 at 4:51
  • $\begingroup$ @user71207 I suppose I could change "Define" to "For", in which case the reader would understand the "is" disappears, but "Define" is also OK because the point is the mechanistic meaning of the symbols, not an attempt to parse everything as English grammar. $\endgroup$
    – J.G.
    Apr 4, 2021 at 6:13
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Let, $$I=\int_{2016}^{3\cdot 2016}\frac{\sqrt[5]{3\cdot 2016-x}}{\sqrt[5]{3\cdot 2016-x}+\sqrt[5]{x- 2016}}\ dx\tag 1$$ Now, using the property of definite integral: $\int_a^bf(x)\ dx=\int_{a}^bf(a+b-x)\ dx$, one should get

\begin{align*} I&=\int_{2016}^{3\cdot 2016}\frac{\sqrt[5]{3\cdot 2016-(4\cdot2016 -x)}}{\sqrt[5]{3\cdot 2016-(4\cdot2016 -x)}+\sqrt[5]{(4\cdot2016 -x)- 2016}}\ dx\\[3ex] I&=\int_{2016}^{3\cdot 2016}\frac{\sqrt[5]{x-2016}}{\sqrt[5]{x-2016}+\sqrt[5]{3\cdot2016 -x}}\ dx\\[3ex] I&=\int_{2016}^{3\cdot 2016}\frac{\sqrt[5]{x-2016}}{\sqrt[5]{3\cdot2016 -x}+\sqrt[5]{x-2016}}\ dx\tag 2\\[6ex] \end{align*}

Now, adding (1) & (2), one should get

\begin{align*} I+I&=\int_{2016}^{3\cdot 2016}\left(\frac{\sqrt[5]{3\cdot 2016-x}}{\sqrt[5]{3\cdot 2016-x}+\sqrt[5]{x- 2016}}+\frac{\sqrt[5]{x-2016}}{\sqrt[5]{3\cdot2016 -x}+\sqrt[5]{x-2016}}\right)\ dx\\[3ex] 2I&=\int_{2016}^{3\cdot 2016}\frac{\sqrt[5]{3\cdot2016 -x}+\sqrt[5]{x-2016}}{\sqrt[5]{3\cdot2016 -x}+\sqrt[5]{x-2016}}\ dx\\[3ex] I&=\frac12\int_{2016}^{3\cdot 2016}\ dx\\[3ex] &=\frac12(3\cdot 2016-2016)\\[3ex] &=\color{red}{2016} \end{align*}

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