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

Question

Evaluate the following integrals:
$\int\limits^1_0\log\Big(\frac{1}{\text{x}}-1\Big)\text{dx}$

✓

Answer

Let $\text{I}=\int\limits^1_0\log\Big(\frac{1}{\text{x}}-1\Big)\text{dx}\ ....(\text{i})$
$=\int\limits^1_0\log\Big(\frac{1}{1-\text{x}}-1\Big)\text{dx}$ $\Big[\text{Using},\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\text{f}(\text{a}-\text{x})\text{dx}\Big]$
$\text{I}=\int\limits^1_0\log\Big(\frac{\text{x}}{\text{x}-1}\Big)\text{dx}\ ....(\text{ii})$
Adding (i) and (ii)
$2\text{I}=\int\limits^1_0\log\Big(\frac{1-\text{x}}{\text{x}}\Big)\log\Big(\frac{1-\text{x}}{\text{x}}\Big)\text{dx}$
$=\int\limits^1_0\log\Big(\frac{1-\text{x}}{\text{x}}\times\frac{\text{x}}{1-\text{x}}\Big)\text{dx}$
$=\int\limits^1_0\log1\text{ dx}$
$=0$
Hence, $\text{I}=0$

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