JEE Math Practice Question

For $$\alpha, \beta, \gamma, \delta \in \mathbb{N}$$, if $$\int\left(\left(\frac{x}{e}\right)^{2 x}+\left(\frac{e}{x}\right)^{2 x}\right) \log _{e} x d x=\frac{1}{\alpha}\left(\frac{x}{e}\right)^{\beta x}-\frac{1}{\gamma}\left(\frac{e}{x}\right)^{\delta x}+C$$ , where $$e=\sum\limits_{n=0}^{\infty} \frac{1}{n !}$$ and $$\mathrm{C}$$ is constant of integration, then $$\alpha+2 \beta+3 \gamma-4 \delta$$ is equal to :

1. A.$$-8$$
2. B.$$-4$$
3. C.1
4. D.4