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$\int_{0}^{1} \log \left(\frac{1}{x}-1\right) d x$ is equal to
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Let $\quad I=\int_{0}^{1} \log \left(\frac{1}{x}-1\right) d x$
$\therefore$
$=\int_{0}^{1} \log \left(\frac{1-x}{x}\right) d x$
$I=\int_{0}^{1} \log \left(\frac{x}{1-x}\right) d x=-1$
$\quad\left[\because \int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\right]$
$2 l=0 \Rightarrow I=0$
$\therefore$
$=\int_{0}^{1} \log \left(\frac{1-x}{x}\right) d x$
$I=\int_{0}^{1} \log \left(\frac{x}{1-x}\right) d x=-1$
$\quad\left[\because \int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\right]$
$2 l=0 \Rightarrow I=0$
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