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

JEE Main · 2023 · 10 Apr 2023 · Shift 2 · Indefinite Integrals

For,

α, β, γ, δ \in N

, if

\int ((\frac{x}{e})^{2 x} + (\frac{e}{x})^{2 x}) lo g_{e} x dx = \frac{1}{α} (\frac{x}{e})^{β x} - \frac{1}{γ} (\frac{e}{x})^{δ x} + C,

Where

e = \sum_{n = 0}^{\infty} \frac{1}{n !}

and C is constant of integration,
then

α + 2 β + 3 γ - 4 δ

is equal to:

Answer: (D)

Put

t = x ln x - x

, so

d t = ln x d x

. Then

(\frac{x}{e})^{2 x} = e^{2 (x l n x - x)} = e^{2 t}

and

(\frac{e}{x})^{2 x} = e^{- 2 t}

, so the integral becomes

\int (e^{2 t} + e^{- 2 t}) d t = \frac{1}{2} e^{2 t} - \frac{1}{2} e^{- 2 t} + C = \frac{1}{2} (\frac{x}{e})^{2 x} - \frac{1}{2} (\frac{e}{x})^{2 x} + C

. Comparing,

α = β = γ = δ = 2

, so

α + 2 β + 3 γ - 4 δ = 2 + 4 + 6 - 8 = 4

. The required value is

4

JEE Main · 2023 · 10 Apr 2023 · Shift 2 · Single correct option