Question

Solve this:

${\int }_0^1\log \left(\frac{1}{x}-1\right)dx$

Answer 1

Dear student
$$\begin{gathered} \mathrm{I}={\int }_0^1\log \left(\frac{1}{\mathrm{x}}-1\right)\mathrm{dx} \\ ={\int }_0^1\log \left(\frac{1-\mathrm{x}}{\mathrm{x}}\right)\mathrm{dx}....\left(1\right) \\ ={\int }_0^1\log \left\{\frac{1-\left(1-\mathrm{x}\right)}{1-\mathrm{x}}\right\}\mathrm{dx}\left[\mathrm{using}{\int }_0^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}={\int }_0^{\mathrm{a}}\mathrm{f}(\mathrm{a}-\mathrm{x})\mathrm{dx}\right] \\ ={\int }_0^1\log \left(\frac{\mathrm{x}}{1-\mathrm{x}}\right)\mathrm{dx}...\left(2\right) \\ \mathrm{Adding}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get} \\ 2\mathrm{I}={\int }_0^1\log \left(\frac{1-\mathrm{x}}{\mathrm{x}}\right)+\log \left(\frac{\mathrm{x}}{1-\mathrm{x}}\right)\mathrm{dx} \\ ={\int }_0^1\log 1\mathrm{dx} \\ =0 \\ \mathrm{So},\mathrm{I}=0 \end{gathered}$$
Regards