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

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(D) Let $I = \int_0^1 \log \left(\frac{1-x}{x}\right) dx$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get: $I = \int_0^1 \log \left(\frac{1-(1-x)}{1-x}\right) dx = \int_0^1 \log \left(\frac{x}{1-x}\right) dx$. $I = \int_0^1 \log \left(\left(\frac{1-x}{x}\right)^{-1}\right) dx = -\int_0^1 \log \left(\frac{1-x}{x}\right) dx$. $I = -I$. $2I = 0$,which implies $I = 0$.

Class 12 Mathematics 7-2.Definite Integral Properties of definite integration

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

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$1/2$
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$1$
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$2$
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$0$

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