The value of ∫e−$$1e2$$ log⁡ xxdx is

∫e−$$1e2$$ log⁡ $$xxdx=$$−∫e−$$11$$ log⁡$$xxdx+$$∫$$1e2$$ log⁡ xxdx (since$$ log⁡x $$0$$ for x∈$$1$$,$$e2$$ $$) $$=$$−∫−$$10$$ $$tdt+$$∫$$02$$ $$tdt=$$−$$t22$$−$$10+t2202=12+2=52$$

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The value of $\int_{e^{- 1}}^{e^{2}} |\frac{\log x}{x}| d x$ is

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Correct option is B

∫e−1e2 log⁡ xxdx=−∫e−11 log⁡xxdx+∫1e2 log⁡ xxdx (since log⁡x<0 for x∈e−1,1 and log⁡x>0 for x∈1,e2 ) =−∫−10 tdt+∫02 tdt=−t22−10+t2202=12+2=52

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The value of ∫e−1e2 log⁡ xxdx is

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The value of $\int_{e^{- 1}}^{e^{2}} |\frac{\log x}{x}| d x$ is