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What is $\int_{e^{-1}}^{\mathrm{e}^{2}}\left|\frac{\ln x}{\mathrm{x}}\right| \mathrm{d} \mathrm{x}$ equal to?
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The correct answer is:
$\frac{5}{2}$
$\int_{x-1}^{e^{2}}\left|\frac{\log x}{x}\right| d x=\int_{s^{-1}}^{e^{0}} \frac{-\log x}{x} d x+\int_{\varepsilon^{0}}^{e^{2}} \frac{\log x}{x} d x$
$=\frac{-1}{2}\left[(\log \mathrm{x})^{2}\right]_{\mathrm{e}}^{\mathrm{e}^{0}}-\frac{1}{2}\left[(\log \mathrm{x})^{2}\right]_{e^{0}}^{\mathrm{c}^{2}}$
$=\frac{-1}{2}\left[0-\left(\log \mathrm{e}^{-1}\right)^{2}\right]+\frac{1}{2}\left[\left(\log \mathrm{e}^{2}\right)^{2}-0\right]$
$=\frac{-1}{2}(-1)+\frac{1}{2}(2)^{2}$
$=\frac{1}{2}+\frac{4}{2}=\frac{5}{2}$
$=\frac{-1}{2}\left[(\log \mathrm{x})^{2}\right]_{\mathrm{e}}^{\mathrm{e}^{0}}-\frac{1}{2}\left[(\log \mathrm{x})^{2}\right]_{e^{0}}^{\mathrm{c}^{2}}$
$=\frac{-1}{2}\left[0-\left(\log \mathrm{e}^{-1}\right)^{2}\right]+\frac{1}{2}\left[\left(\log \mathrm{e}^{2}\right)^{2}-0\right]$
$=\frac{-1}{2}(-1)+\frac{1}{2}(2)^{2}$
$=\frac{1}{2}+\frac{4}{2}=\frac{5}{2}$
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