$\underset{0}{\overset{1}{\int }}{(1-x^7)}^{\frac{1}{4}}dx-\underset{0}{\overset{1}{\int }}{(1-x^4)}^{\frac{1}{7}}dx$ is equal to 

$\underset{0}{\overset{1}{\int }}{(1-x^7)}^{\frac{1}{4}}dx-\underset{0}{\overset{1}{\int }}{(1-x^4)}^{\frac{1}{7}}dx$

 $y=f\left(x\right)={(1-x^7)}^{\frac{1}{4}}\Rightarrow y^4=1-x^7\Rightarrow x={(1-y^4)}^{\frac{1}{7}}$

Hence function are inverse of each other
Hence  $I=\underset{0}{\overset{1}{\int }}(f\left(x\right)+xf\text{'}cx)dx=xf{\left(x\right)]}_0^1=f\left(1\right)=0$