If {I_1} = int_e^{{e^2}} frac{dx}{log x} and | vedclass

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(A) To evaluate ${I_1} = \int_e^{{e^2}} \frac{dx}{\log x}$,we use the substitution method.Let $\log x = u$. Then $x = e^u$,which implies $dx = e^u du$

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${I_1} = {I_2}$

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(A) To evaluate ${I_1} = \int_e^{{e^2}} \frac{dx}{\log x}$,we use the substitution method.Let $\log x = u$. Then $x = e^u$,which implies $dx = e^u du$.When $x = e$,$u = \log e = 1$.When $x = e^2$,$u = \log e^2 = 2$.Substituting these into the integral ${I_1}$,we get:${I_1} = \int_1^2 \frac{e^u}{u} du$.Since the variable of integration is a dummy variable,we can replace $u$ with $x$:${I_1} = \int_1^2 \frac{e^x}{x} dx$.Comparing this with ${I_2} = \int_1^2 \frac{e^x}{x} dx$,we see that ${I_1} = {I_2}$.

If ${I_1} = \int_e^{{e^2}} \frac{dx}{\log x}$ and ${I_2} = \int_1^2 \frac{e^x}{x} dx$,then

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