intlimits_2^4 {left[ {{{log }_x}2 - frac{{{{l | vedclass

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(A) Let $I = \int\limits_2^4 {\left[ {{{\log }_x}2 - \frac{{{{\left( {{{\log }_x}2} \right)}^2}}}{{\ln 2}}} \right]} dx$. Using the change of base for

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(A) Let $I = \int\limits_2^4 {\left[ {{{\log }_x}2 - \frac{{{{\left( {{{\log }_x}2} \right)}^2}}}{{\ln 2}}} \right]} dx$. Using the change of base formula,$\log_x 2 = \frac{\ln 2}{\ln x}$. Substituting this into the integral: $I = \int\limits_2^4 {\left( {\frac{{\ln 2}}{{\ln x}} - \frac{{{{(\ln 2)}^2}}}{{{{(\ln x)}^2} \cdot \ln 2}}} \right)} dx = \ln 2 \int\limits_2^4 {\left( {\frac{1}{{\ln x}} - \frac{1}{{{{(\ln x)}^2}}}} \right)} dx$. Let $f(x) = \frac{x}{\ln x}$. Then $f'(x) = \frac{\ln x - x(1/x)}{{(\ln x)^2}} = \frac{\ln x - 1}{{(\ln x)^2}} = \frac{1}{\ln x} - \frac{1}{{(\ln x)^2}}$. Thus,$I = \ln 2 \left[ {\frac{x}{{\ln x}}} \right]_2^4$. $I = \ln 2 \left( {\frac{4}{{\ln 4}} - \frac{2}{{\ln 2}}} \right) = \ln 2 \left( {\frac{4}{{2\ln 2}} - \frac{2}{{\ln 2}}} \right) = \ln 2 \left( {\frac{2}{{\ln 2}} - \frac{2}{{\ln 2}}} \right) = 0$.

$\int\limits_2^4 {\left[ {{{\log }_x}2 - \frac{{{{\left( {{{\log }_x}2} \right)}^2}}}{{\ln 2}}} \right]} dx =$

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