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

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(D) माना $I = \int_0^1 \log \left(\frac{1-x}{x}\right) dx$. गुणधर्म $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ का उपयोग करने पर,हमें प्राप्त होता है: $I

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(D) माना $I = \int_0^1 \log \left(\frac{1-x}{x}\right) dx$. गुणधर्म $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ का उपयोग करने पर,हमें प्राप्त होता है: $I = \int_0^1 \log \left(\frac{1-(1-x)}{1-x}\right) dx = \int_0^1 \log \left(\frac{x}{1-x}\right) dx$. $I = \int_0^1 \log \left(\left(\frac{1-x}{x}\right)^{-1}\right) dx = -\int_0^1 \log \left(\frac{1-x}{x}\right) dx$. $I = -I$. $2I = 0$,जिसका अर्थ है कि $I = 0$.

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

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