Question

The value of ${\int }_{-1}^1\log \left(\frac{x-1}{x+1}\right)dx$ is

• A. $1$
• B. $2$
• C. $0$
• D. $4$

Answer 1

The correct option is C $0$

Let $I={\int }_{-1}^1\log \left(\frac{x-1}{x+1}\right)dx$

Again, let $f\left(x\right)=\log \left(\frac{x-1}{x+1}\right)$

$f(-x)=\log \left(\frac{-x-1}{-x+1}\right)$

$=\log \left(\frac{1+x}{x-1}\right)$

$=-\log \left(\frac{x-1}{x+1}\right)$

$=-f\left(x\right)$

$\Rightarrow f\left(x\right)$ is an odd function.

$\therefore I=0$