Assertion (A): int_2^e left(frac{1}{log_e x} | vedclass

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(A) Assertion: Let $I = \int_2^e \left(\frac{1}{\log_e x} - \frac{1}{(\log_e x)^2}\right) dx$. Let $\log_e x = y$,then $x = e^y$ and $dx = e^y dy$. Wh

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$A$ and $R$ are true,$R$ is the correct explanation to $A$.

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(A) Assertion: Let $I = \int_2^e \left(\frac{1}{\log_e x} - \frac{1}{(\log_e x)^2}\right) dx$. Let $\log_e x = y$,then $x = e^y$ and $dx = e^y dy$. When $x = 2$,$y = \log_e 2$. When $x = e$,$y = 1$. Substituting these into the integral: $I = \int_{\log_e 2}^1 e^y \left(\frac{1}{y} - \frac{1}{y^2}\right) dy$. Using the formula $\int e^y (f(y) + f'(y)) dy = e^y f(y) + C$,where $f(y) = \frac{1}{y}$ and $f'(y) = -\frac{1}{y^2}$. $I = \left[ e^y \cdot \frac{1}{y} \right]_{\log_e 2}^1 = \left( e^1 \cdot \frac{1}{1} \right) - \left( e^{\log_e 2} \cdot \frac{1}{\log_e 2} \right) = e - \frac{2}{\log_e 2} = e - 2 \log_2 e$. The Reason $(R)$ provides the standard formula $\int_a^b e^x (f(x) + f'(x)) dx = [e^x f(x)]_a^b$,which is correct. Thus,both Assertion and Reason are true,and the Reason is the correct explanation of the Assertion.

Assertion $(A)$: $\int_2^e \left(\frac{1}{\log_e x} - \frac{1}{(\log_e x)^2}\right) dx = e - 2 \log_2 e$
Reason $(R)$: $\int_a^b e^x (f(x) + f'(x)) dx = e^b f(b) - e^a f(a)$

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Assertion $(A)$: $\int_2^e \left(\frac{1}{\log_e x} - \frac{1}{(\log_e x)^2}\right) dx = e - 2 \log_2 e$ Reason $(R)$: $\int_a^b e^x (f(x) + f'(x)) dx = e^b f(b) - e^a f(a)$