For α , β , γ , δ ∈ ℕ , if ∫ x e 2 x + e x 2 x log e x d x = 1 α x e β x - 1 γ e x δ x + C , where e = ∑ n = 0 ∞ 1 n ! and C is constant of integration, then α + 2 β + 3 γ - 4 δ is equal to

Step-by-step solution

Given, ∫ x e 2 x + e x 2 x ln x d x = 1 α x e β x - 1 γ e x δ x + C Now let I = ∫ x e 2 x + e x 2 x ln x d x Now let x e 2 x = t ⇒ 2 x ln x - 1 = ln t ⇒ ln x d x = 1 2 t d t So, I = 1 2 ∫ t + 1 t d t t ⇒ I = 1 2 ∫ 1 + 1 t 2 d t ⇒ I = 1 2 t - 1 t + c ⇒ I = 1 2 x e 2 x - e x 2 x + C Now on comparing with I = 1 α x e β x - 1 γ e x δ x + C , we get α = 2 ,   β = 2 , γ = 2 , δ = 2 Hence, α + 2 &#94