Question

What is the integration of $\log \left(\frac{1-x}{x}\right)$ with limit $0$ to $1$?

Answer 1

Find the integration of $\log \left(\frac{1-x}{x}\right)$ with limit $0$ to $1$.

$\begin{array}{rcl}{\int }_0^1\log \left(\frac{1-x}{x}\right)dx& =& {\left[\log \left(\frac{1-x}{x}\right).x-\int x\left(\frac{x}{1-x}\right)\left(\frac{-1}{x^2}-0\right)dx\right]}_0^1\left[\because Useintegrationbypartsi.e.\int uvdx=u\int vdx-\int \left[\left(\int vdx\right)\left(\frac{du}{dx}\right)dx\right]\right]\\ & \Rightarrow & {\int }_0^1\log \left(\frac{1-x}{x}\right)dx={\left[\log \left(\frac{1-x}{x}\right).x+\int \left(\frac{1}{1-x}\right)dx\right]}_0^1\\ & \Rightarrow & {\int }_0^1\log \left(\frac{1-x}{x}\right)dx={\left[\log \left(\frac{1-x}{x}\right).x-\log \left(1-x\right)\right]}_0^1\\ & \Rightarrow & {\int }_0^1\log \left(\frac{1-x}{x}\right)dx=\left[\log \left(\frac{1-1}{1}\right).1-\log \left(1-1\right)\right]-\left[\log \left(\frac{1-0}{0}\right).0-\log \left(1-0\right)\right]\\ & \Rightarrow & {\int }_0^1\log \left(\frac{1-x}{x}\right)dx=0\end{array}$Hence, the required integral is ${\int }_0^1\log \left(\frac{1-x}{x}\right)\mathrm{dx}=0$