For, α, β, γ, δ ∈ \(\mathbb{N},\) if \(∫\left((\frac{x}{e})^{2x}+(\frac{e}{x})^{2x}\right)log_e\,xdx=\) \(\frac{1}{\alpha}(\frac{x}{e})^{βx}-\frac{1}{γ}(\frac{e}{x})^{δx}+c,\) Where \(e=\displaystyle\sum_{n=10}^{∞}\frac{1}{n!}\) and C is constant of integration, then α + 2β + 3γ - 4δ is equal to