Why is $\int\limits_0^1 (1-x^7)^{1/5} - (1-x^5)^{1/7} dx=0$?

Note that if

$$ y = \left(1 - x^7\right)^{1/5} $$

then

$$ \left(1 - y^5\right)^{1/7} = x $$

This means $(1-x^7)^{1/5}$ is the inverse function of $(1-x^5)^{1/7}$. In the graph, one will be the same as the other when reflected along the diagonal line y = x.

Also, both functions

1. share the same range [0, 1] and domain [0, 1] and
2. monotonically decreasing,

Therefore, the area under the graph in [0, 1] will be the same for both functions:

$$ \int_0^1 \left(1-x^7\right)^{1/5} dx = \int_0^1 \left(1-y^5\right)^{1/7} dy $$

Grouping the two integrals yield the equation in the title.

$\int_0^1(1-x^m)^{(1/n)}dx=(m+n)\Gamma(1/m)\Gamma(1/n)/\Gamma(1/m+1/n)$ is symmetric in $m, n$.