If int_2^{e}left[frac{1}{log x}-frac{1}{(log | vedclass

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(B) We are given the integral $I = \int_2^{e}\left[\frac{1}{\log x}-\frac{1}{(\log x)^2}\right] dx$. Let $\log x = t$,then $x = e^t$ and $dx = e^t dt$

Vedclass · Accepted Answer

$a=e, b=-2$

Vedclass · Answer

(B) We are given the integral $I = \int_2^{e}\left[\frac{1}{\log x}-\frac{1}{(\log x)^2}\right] dx$. Let $\log x = t$,then $x = e^t$ and $dx = e^t dt$. When $x = 2$,$t = \log 2$. When $x = e$,$t = 1$. Substituting these into the integral: $I = \int_{\log 2}^{1} \left(\frac{1}{t} - \frac{1}{t^2}\right) e^t dt$. We know that $\int e^t [f(t) + f'(t)] dt = e^t f(t) + C$. Here,$f(t) = \frac{1}{t}$ and $f'(t) = -\frac{1}{t^2}$. Therefore,$I = \left[ e^t \cdot \frac{1}{t} \right]_{\log 2}^{1}$. $I = \left( e^1 \cdot \frac{1}{1} \right) - \left( e^{\log 2} \cdot \frac{1}{\log 2} \right)$. $I = e - \frac{2}{\log 2}$. Comparing this with $a + \frac{b}{\log 2}$,we get $a = e$ and $b = -2$.

If $\int_2^{e}\left[\frac{1}{\log x}-\frac{1}{(\log x)^2}\right] dx = a+\frac{b}{\log 2}$,then:

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