Rajasthan Boardहिन्दी माध्यमकक्षा 12 साइन्सगणितसमाकलन1 Mark
Question
$\int_0^1 \log \left(\frac{1}{x}-1\right) d x$ का मान है-
✓
Answer
(A) $0$ माना $\quad I =\int_0^1 \log \left(\frac{1}{x}-1\right) d x=\int_0^1 \log \left(\frac{1-x}{x}\right) d x \ldots(1)$ $I=\int_0^1 \log \left(\frac{1-x}{x}\right) d x$ $=\int_0^1 \log \left(\frac{1-(1-x)}{(1-x)}\right) d x$ गुणधर्म $P _5$ से $I=\int_0^1 \log \left(\frac{1-1+x}{(1-x)}\right) d x=\int_0^1 \log \left(\frac{x}{(1-x)}\right) d x$ $I=-\int_0^1 \log \left(\frac{1-x}{x}\right) d x$ $\begin{aligned} I & =-I \\ 2 I & =0 \therefore I=0\end{aligned}$ अत: सही विकल्प (A) है।
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