For alpha, beta, gamma, delta in mathbb{N},if | vedclass

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(D) We have the integral $I = \int \left( \left( \frac{x}{e} \right)^{2x} + \left( \frac{e}{x} \right)^{2x} \right) \ln x \, dx$. Note that $\left( \f

Vedclass · Accepted Answer

$4$

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(D) We have the integral $I = \int \left( \left( \frac{x}{e} \right)^{2x} + \left( \frac{e}{x} \right)^{2x} \right) \ln x \, dx$. Note that $\left( \frac{x}{e} \right)^{2x} = e^{2x \ln(x/e)} = e^{2x(\ln x - 1)} = e^{2(x \ln x - x)}$. Similarly,$\left( \frac{e}{x} \right)^{2x} = e^{-2(x \ln x - x)}$. Let $t = x \ln x - x$. Then $dt = (\ln x + x \cdot \frac{1}{x} - 1) \, dx = \ln x \, dx$. Substituting these into the integral,we get $I = \int (e^{2t} + e^{-2t}) \, dt$. Integrating,we obtain $I = \frac{e^{2t}}{2} - \frac{e^{-2t}}{2} + C$. Substituting back $t = x \ln x - x$,we get $I = \frac{1}{2} \left( \frac{x}{e} \right)^{2x} - \frac{1}{2} \left( \frac{e}{x} \right)^{2x} + C$. Comparing this with the given form,we have $\alpha = 2, \beta = 2, \gamma = 2, \delta = 2$. Thus,$\alpha + 2\beta + 3\gamma - 4\delta = 2 + 2(2) + 3(2) - 4(2) = 2 + 4 + 6 - 8 = 4$.

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