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As a Happy New Year card for Stack Exchange communities goes this problem.

Prove the equality $${{\large\int_{-2017}^{2017}\left(\frac{\large\displaystyle\sqrt[2017]{x-2017}}{\space\large\displaystyle\sqrt[2017]{x-2017}+\large\displaystyle\sqrt[2017]{x+2017}\space }\right)dx=2017}}$$

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    $\begingroup$ Dear @Piquito: How do you come up with such integrals? Last year also you asked such question which can be found [here] (math.stackexchange.com/questions/1595361/…). $\endgroup$ – Arpit Kansal Dec 29 '16 at 11:58
  • $\begingroup$ @Arpit Kansal: Yes but this integral is not of the same form. This is very easy to solve despite appearances. Best regards. $\endgroup$ – Piquito Dec 29 '16 at 12:05
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    $\begingroup$ Dear @Piquito: I know its easy but i was just interested how do you construct such integrals? $\endgroup$ – Arpit Kansal Dec 29 '16 at 12:15
  • $\begingroup$ I am also interested. $\endgroup$ – Kanwaljit Singh Dec 29 '16 at 12:16
  • $\begingroup$ See the used property in Olivier Oloa's answer. Notice that the exponent can be arbitrary, not just $\frac{1}{2017}$. Also you can use the quite simple change of variable $x=-t$ $\endgroup$ – Piquito Dec 29 '16 at 14:19
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One may use the property $$ \int_a^bf(x)\ dx=\int_{a}^bf(a+b-x)\ dx $$ applied to $$ f(x)=\frac{\sqrt[2017]{x-2017}}{\sqrt[2017]{x-2017}+\sqrt[2017]{x+2017}\space }, \quad a=-2017,\quad b=2017, $$ giving $$ \begin{align} I=\int_{-2017}^{2017}\frac{\sqrt[2017]{x-2017}}{\sqrt[2017]{x-2017}+\sqrt[2017]{x+2017}}\ dx&=\int_{-2017}^{2017}\frac{\sqrt[2017]{-x-2017}}{\sqrt[2017]{-x-2017}+\sqrt[2017]{-x+2017}}\ dx \\\\&=\int_{-2017}^{2017}\frac{\color{red}{-}\:\sqrt[2017]{x+2017}}{\color{red}{-}\:\sqrt[2017]{x+2017}\color{red}{-}\:\sqrt[2017]{x-2017}}\ dx \\\\&=\int_{-2017}^{2017}\frac{\sqrt[2017]{x+2017}}{\sqrt[2017]{x+2017}+\sqrt[2017]{x-2017}}\ dx \end{align} $$ thus $$ 2I=I+I=\int_{-2017}^{2017}\frac{\sqrt[2017]{x-2017}+\sqrt[2017]{x+2017}}{\sqrt[2017]{x-2017}+\sqrt[2017]{x+2017}}\ dx=\int_{-2017}^{2017}\ dx=2\cdot 2017 $$ that is $$ I=2017. $$

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    $\begingroup$ +1,last year question was also the direct application of same formula!(See my comment) $\endgroup$ – Arpit Kansal Dec 29 '16 at 12:37
  • $\begingroup$ @Olivier: See the comment for hypergeometric above $\endgroup$ – Piquito Dec 30 '16 at 15:23
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$$\text{If }J=\int_a^b\frac{g(x)}{g(x)+g(a+b-x)}dx,\ J=\int_a^b\frac{g(a+b-x)}{g(x)+g(a+b-x)}dx$$

$$\implies J+J=\int_a^b dx$$ provided $g(x)+g(a+b-x)\ne0$

If $\displaystyle f(x)=\sqrt[2n+1]{x-2017}, f(2017-2017-x)=-\ \sqrt[2n+1]{x+2017}$

For An integral for the New Year,

if $\displaystyle g(x)=\sqrt[2n+1]{3\cdot2016-x},\ g(3\cdot2016+2016-x)=\ ?$

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