If \(∫\left((\frac{x}{e})^{4x}+(\frac{e}{x})^{4x}\right)\ln xdx\) = \(\frac{1}{\alpha}(\frac{x}{e})^{βx}-\frac{1}{γ}(\frac{e}{x})^{δx}+c\), then the value of αβ + γδ is