Oh Oh I know this one!

$\int_{0}^{1}((1-x^{9})^{\frac{1}{5}}-(1-x^{5})^{\frac{1}{9}})dx=\int_{0}^{1}(1-x^{9})^{\frac{1}{5}}dx-\int_{0}^{1}(1-x^{5})^{\frac{1}{9}}$

Notice how this is just the area under $x^{9}+x^{5}=1$

AREA - AREA = 0

$\begin{aligned} \int_0^1 {\left(\left(1-x^9\right)^{\frac{1}{5}}-\left(1-x^5\right)^{\frac{1}{9}}\right)dx} & = \int_0^1 {\left(1-x^9\right)^{\frac{1}{5}}dx} - \int_0^1 {\left(1-x^5\right)^{\frac{1}{9}}dx} \end{aligned}$

Let $\space t = x^9\quad \Rightarrow dt = 9x^8 dx \quad \Rightarrow dx = \dfrac{1}{9}t^{-\frac{8}{9}} dt$

$\begin{aligned} \Rightarrow \int_0^1 {\left(1-x^9\right)^{\frac{1}{5}}dx} & = \frac{1}{9} \int_0^1 { t^{-\frac{8}{9}} \left(1-t\right)^{\frac{1}{5}} dt} \\ & = \frac{1}{9} \int_0^1 { t^{\frac{1}{9} - 1} \left(1-t \right)^{\frac{6}{5} - 1} dt} \\ & = \frac{1}{9} \color{#3D99F6}{B \left(\frac{1}{9}, \frac{6}{5} \right) \quad [B(m,n) = \text{Beta function}]} \\ & = \color{#3D99F6} {\frac {\Gamma \left(\frac{1}{9} \right) \Gamma \left( \frac{6}{5} \right)} {9\Gamma \left(\frac{59}{45} \right)}\quad \space [\Gamma (p) = \text{Gamma function}]} \\ & = \frac {\frac{1}{5}\Gamma \left(\frac{1}{9} \right) \Gamma \left( \frac{1}{5} \right)} {9\times \frac{14}{45} \Gamma \left(\frac{14}{45} \right)} = \frac {\Gamma \left(\frac{1}{9} \right) \Gamma \left( \frac{1}{5} \right)} {14\Gamma \left(\frac{14}{45} \right)} \end{aligned}$

Similarly,

$\begin{aligned} \Rightarrow \int_0^1 {\left(1-x^5\right)^{\frac{1}{9}}dx} & = \frac{1}{5} \int_0^1 { t^{-\frac{4}{5}} \left(1-t\right)^{\frac{1}{9}} dt} \\ & = \frac{1}{5} \int_0^1 { t^{\frac{1}{5} - 1} \left(1-t \right)^{\frac{10}{9} - 1} dt} \\ & = \frac{1}{5} B \left(\frac{1}{5}, \frac{10}{9} \right) = \frac {\Gamma \left(\frac{1}{5} \right) \Gamma \left( \frac{10}{9} \right)} {5\Gamma \left(\frac{59}{45} \right)} = \frac {\frac{1}{9}\Gamma \left(\frac{1}{5} \right) \Gamma \left( \frac{1}{9} \right)} {5\times \frac{14}{45} \Gamma \left(\frac{14}{45} \right)} \\ & = \frac {\Gamma \left(\frac{1}{5} \right) \Gamma \left( \frac{1}{9} \right)} {14\Gamma \left(\frac{14}{45} \right)} = \int_0^1 {\left(1-x^9\right)^{\frac{1}{5}}dx} \end{aligned}$

$\Rightarrow \displaystyle \int_0^1 {\left(\left(1-x^9\right)^{\frac{1}{5}}-\left(1-x^5\right)^{\frac{1}{9}}\right)dx} = \boxed{0}$

Let's start from change x to u such that

$(1-x^{9})^{\dfrac{1}{5}}=u$ then $x=(1-u^{5})^{\dfrac{1}{9}}$

substitute this into the first integral

$\displaystyle \int_{0}^{1} (1-x^{9})^{\dfrac{1}{5}} dx = \int_{1}^{0} u d(1-u^{5})^{\dfrac{1}{9}}$

$\displaystyle - \int_{0}^{1} u d (1-u^{5})^{\dfrac{1}{9}} = - u\cdot (1-u^{5})^{\dfrac{1}{9}} + \int_{0}^{1} (1-u^{5})^{\dfrac{1}{9}} du$

the latter term vanish after combine with the latter integral.So we have

$\displaystyle - u\cdot (1-u^{5})^{\dfrac{1}{9}}$ from $0$ to $1$ which is 0.