∫1/ee2ln xxdx =∫1/e1ln xx + ∫1e2ln xxdx =∫1/e1In xxdx+∫1e2In xxdx =-(In x)221/e1+In x221e2 =12(1-0)+12(4-0) =52

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$\int_{1 / e}^{e^{2}} |\frac{\ln x}{x}| d x =$

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Detailed Solution

$\int_{1 / e}^{e^{2}} |\frac{\ln x}{x}| dx = \int_{1 / e}^{1} |\frac{\ln x}{x}| + \int_{1}^{e^{2}} |\frac{\ln x}{x}| dx = \int_{1 / e}^{1} \frac{I n x}{x} dx + \int_{1}^{e^{2}} \frac{In x}{x} dx = {(\frac{- {(In x)}^{2}}{2})}_{1 / e}^{1} + {(\frac{{(In x)}^{2}}{2})}_{1}^{e^{2}} = \frac{1}{2} (1 - 0) + \frac{1}{2} (4 - 0) = \frac{5}{2}$