An integral for the New Year 2016

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$$I=\frac12\int_k^{3k}dx=k.$$

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*}