Evaluate the following integrals : _0^1 (1 x -1 ) d x - Vidyadip

Question

Evaluate the following integrals : $\int_0^1 \log \left(\frac{1}{x}-1\right) d x$

✓

Answer

Let $I=\int_0^1 \log \left(\frac{1}{x}-1\right) d x=\int_0^1 \log \left(\frac{1-x}{x}\right) d x$
We use the property, $\int_0^a f(x) d x=\int_0^a f(a-x) d x$
$
\begin{aligned}
\therefore 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 \\
& =\int_0^1-\log \left(\frac{1-x}{x}\right) d x=-\int_0^1 \log \left(\frac{1-x}{x}\right) d x
\end{aligned}
$
$
\begin{aligned}
& \therefore I=-I \\
& \therefore 2 I=0 \quad \therefore I=0
\end{aligned}
$
Hence, $\int_0^1 \log \left(\frac{1}{x}-1\right) d x=0$.

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