int_{0}^{1} log left(frac{1}{x}-1right) d x i | vedclass

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(B) Let $I = \int_{0}^{1} \log \left(\frac{1-x}{x}\right) d x$. Using the property $\int_{a}^{b} f(x) d x = \int_{a}^{b} f(a+b-x) d x$, we get: $I = \

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(B) Let $I = \int_{0}^{1} \log \left(\frac{1-x}{x}\right) d x$. Using the property $\int_{a}^{b} f(x) d x = \int_{a}^{b} f(a+b-x) d x$, we get: $I = \int_{0}^{1} \log \left(\frac{1-(1-x)}{1-x}\right) d x$ $I = \int_{0}^{1} \log \left(\frac{x}{1-x}\right) d x$ $I = \int_{0}^{1} \log \left(\left(\frac{1-x}{x}\right)^{-1}\right) d x$ $I = -\int_{0}^{1} \log \left(\frac{1-x}{x}\right) d x$ $I = -I$ $2I = 0 \implies I = 0$.

$\int_{0}^{1} \log \left(\frac{1}{x}-1\right) d x$ is equal to

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