MathematicsHard64×since 2004Q4055For α,β,γ,δ∈N\alpha, \beta, \gamma, \delta \in \mathbb{N}α,β,γ,δ∈N, if ∫((xe)2x+(ex)2x)logexdx=1α(xe)βx−1γ(ex)δx+C\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∫((ex)2x+(xe)2x)logexdx=α1(ex)βx−γ1(xe)δx+C , where e=\sum_\limits{n=0}^{\infty} \frac{1}{n !} and C\mathrm{C}C is constant of integration, then α+2β+3γ−4δ\alpha+2 \beta+3 \gamma-4 \deltaα+2β+3γ−4δ is equal to :A−8-8−8B−4-4−4C1D4Check answerSkip