# An integral with $2017$

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

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

• 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/…). – Arpit Kansal Dec 29 '16 at 11:58
• @Arpit Kansal: Yes but this integral is not of the same form. This is very easy to solve despite appearances. Best regards. – Piquito Dec 29 '16 at 12:05
• Dear @Piquito: I know its easy but i was just interested how do you construct such integrals? – Arpit Kansal Dec 29 '16 at 12:15
• I am also interested. – Kanwaljit Singh Dec 29 '16 at 12:16
• 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 $x=-t$ – Piquito Dec 29 '16 at 14:19

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 $$I=2017.$$
$$\text{If }J=\int_a^b\frac{g(x)}{g(x)+g(a+b-x)}dx,\ J=\int_a^b\frac{g(a+b-x)}{g(x)+g(a+b-x)}dx$$
$$\implies J+J=\int_a^b dx$$ provided $g(x)+g(a+b-x)\ne0$
If $\displaystyle f(x)=\sqrt[2n+1]{x-2017}, f(2017-2017-x)=-\ \sqrt[2n+1]{x+2017}$
if $\displaystyle g(x)=\sqrt[2n+1]{3\cdot2016-x},\ g(3\cdot2016+2016-x)=\ ?$