_0^1 (1 x -1 ) d x is equal to : - Vidyadip

Question

$\int_0^1 \log \left(\frac{1}{x}-1\right) d x$ is equal to :

✓

Answer

(A)Let
$
\begin{aligned}
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 \text { From Property } P_5
\end{aligned}
$
$
\begin{aligned}
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 \\
I & =-I \\
2 I & =0 \therefore I=0
\end{aligned}
$
Hence option (A) is correct.

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