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

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Question

(A) Let $I = \int_2^e {\left[ {\frac{1}{{\log x}} - \frac{1}{{{{(\log x)}^2}}}} \right]} \,dx$. Using integration by parts on the first term $\int \fr

Vedclass · Accepted Answer

$\alpha = e, \beta = -2$

Vedclass · Answer

(A) Let $I = \int_2^e {\left[ {\frac{1}{{\log x}} - \frac{1}{{{{(\log x)}^2}}}} \right]} \,dx$. Using integration by parts on the first term $\int \frac{1}{\log x} dx$: Let $u = \frac{1}{\log x}$ and $dv = dx$. Then $du = -\frac{1}{x(\log x)^2} dx$ and $v = x$. So,$\int \frac{1}{\log x} dx = \frac{x}{\log x} - \int x \left( -\frac{1}{x(\log x)^2} \right) dx = \frac{x}{\log x} + \int \frac{1}{(\log x)^2} dx$. Substituting this back into the integral $I$: $I = \left[ \frac{x}{\log x} + \int \frac{1}{(\log x)^2} dx \right]_2^e - \int_2^e \frac{1}{(\log x)^2} dx$. $I = \left[ \frac{x}{\log x} \right]_2^e = \frac{e}{\log e} - \frac{2}{\log 2} = e - \frac{2}{\log 2}$. Comparing this with $\alpha + \frac{\beta}{\log 2}$,we get $\alpha = e$ and $\beta = -2$.

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

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