Question
Evaluate the following integrals:$\int\limits^{\text{e}^2}_\text{e}\frac{1}{\text{x}\log\text{x}}\text{ dx}$

Answer

$\int\limits^{\text{e}^2}_\text{e}\frac{1}{\text{x}\log\text{x}}\text{ dx}$$=\int\limits^{\text{e}^2}_\text{e}\frac{\frac{1}{\text{x}}}{\log\text{x}}\text{ dx}$
$=\log\big[(\log{\text{x}})\big]^{\text{e}^2}_\text{e}$ $\Big[\int\frac{\text{f}'(\text{x})}{\text{f(x)}}\text{ dx}=\log\text{f(x)}+\text{C}\Big]$
$=\log\big(\log\text{e}^2\big)-\log(\log\text{e})$
$=\log(2\log\text{e})-\log(\log\text{e})$
$=\log2-\log1$ $(\log\text{e}=1)$
$=\log2-0$
$=\log2$

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