int_0^1 log left(frac{1}{x}-1right) d x is | vedclass

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(B) Let $I = \int_0^1 \log \left(\frac{1-x}{x}\right) d x$. Using the property $\int_0^a f(x) d x = \int_0^a f(a-x) d x$,we get: $I = \int_0^1 \log \l

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(B) Let $I = \int_0^1 \log \left(\frac{1-x}{x}ight) d x$. Using the property $\int_0^a f(x) d x = \int_0^a f(a-x) d x$,we get: $I = \int_0^1 \log \left(\frac{1-(1-x)}{1-x}ight) d x$ $I = \int_0^1 \log \left(\frac{x}{1-x}ight) d x$ Adding the two expressions for $I$: $2I = \int_0^1 \left[ \log \left(\frac{1-x}{x}ight) + \log \left(\frac{x}{1-x}ight) ight] d x$ $2I = \int_0^1 \log \left( \frac{1-x}{x} 	imes \frac{x}{1-x} ight) d x$ $2I = \int_0^1 \log(1) d x$ Since $\log(1) = 0$,we have $2I = 0$,which implies $I = 0$.

$\int_0^1 \log \left(\frac{1}{x}-1\right) d x$ is

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