यदि \(∫\left((\frac{x}{e})^{4x}+(\frac{e}{x})^{4x}\right)\ln xdx\) = \(\frac{1}{\alpha}(\frac{x}{e})^{βx}-\frac{1}{γ}(\frac{e}{x})^{δx}+c\) है, तो αβ + γδ का मान है:

1
16
2
32
3
44
4
64

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