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प्रश्न
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उत्तर
\[Let\ I = \int_0^1 \log\left( \frac{1}{x} - 1 \right) d x ...............(1)\]
\[ = \int_0^1 \log\left( \frac{1}{1 - x} - 1 \right) d x ...............\left[\text{Using }\int_0^a f(x) dx = \int_0^a f(a - x) dx \right]\]
\[ I = \int_0^1 \log\left( \frac{x}{1 - x} \right) dx ...............(2)\]
\[\text{Adding (1) and (2)}\]
\[2I = \int_0^1 \log\left( \frac{1 - x}{x} \right) + \log\left( \frac{x}{1 - x} \right) dx\]
\[ = \int_0^1 \log\left( \frac{1 - x}{x} \times \frac{x}{1 - x} \right) dx\]
\[ = \int_0^1 \log1 dx \]
\[ = 0\]
\[Hence\ I = 0\]
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