Question

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

Answer 1

$\begin{array}{c}{\int }_0^1\log \left(\frac{1-x}{x}\right)\\ \Rightarrow \log \left(\frac{1-x}{x}\right)\cdot {\int }_0^{1}dx-{\int }_0^1\left[\frac{d\left[\log \left(\frac{1-x}{x}\right)\right]}{dx}fdx\right]dx\\ \Rightarrow {\left[x\log \left(\frac{1-x}{x}\right)\right]}_0^1-{\int }_0^1\frac{x}{\left(1-x\right)}\left(\frac{-x-1+x}{x^2}\right)\cdot xdx\\ \Rightarrow {\left[x\log \left(\frac{1-x}{x}\right)\right]}_0^1+{\int }_0^1\frac{1}{1-x}dx\\ \Rightarrow {\left[x\log \left(\frac{1-x}{x}\right)+\log \left(1-x\right)\right]}_0^1+c\\ \Rightarrow {\left[\log {\left(\frac{1-x}{x}\right)}^x-\log \left(1-x\right)\right]}_0^1\\ \Rightarrow {\left[\log \left(\frac{{\left(1-x\right)}^2}{x{\cdot }^x\left(1-x\right)}\right)\right]}_0^1\\ \Rightarrow {\left[\log \left(\frac{{\left(1-x\right)}^{\left(x-1\right)}}{x^x}\right)\right]}_0^1=\log o-\log 0=0Ans.\end{array}$