Question

If $f(x)={{\log }_{e}}\,(1+x)-{{\log }_{e}}(1-x),$ then the value of $\int_{-1/2}^{1/2}{f(x)\,\,\,dx}$ equals to

• A. $0$
• B. $1$
• C. $\frac{1}{2}$
• D. $-\frac{1}{2}$

Answer 1

Answer: (A)

$0$

Answer 2

Given, $f(x)={{\log }_{e}}(1+x)-{{\log }_{e}}(1-x)$
$\therefore$ $f(-x)={{\log }_{e}}(1-x)-{{\log }_{e}}(1+x)$
$=-[{{\log }_{e}}(1+x)-{{\log }_{e}}(1-x)]$
$=-f(x)$
$\therefore$ $f(x)$ is an odd function.
$\therefore$ $\int_{-1/2}^{1/2}{f(x)\,dx=0}$