JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

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Analyzing +4 | -1

For

\alpha, \beta, \gamma, \delta \in \mathbb{N}

, if

\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)^{8 x}+C

, where

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

and

C

is constant of integration, then

\alpha+2 \beta+3 \gamma-4 \delta

is equal to

View answer and explanation

Correct option: B

Option B (4) is correct because after substituting

t = (x/e)^{2x}

, the integral simplifies and can be easily evaluated, leading to the values of

\alpha, \beta, \gamma, \delta

and thus the final result.

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