Question

lf $x>0$ then the value of ${\tanh }^{-1}\left(\frac{x^2-1}{x^2+1}\right)$

• A. ${\log }_e\left(2x\right)$
• B. ${\log }_{e}x$
• C. ${\log }_e\left(3x\right)$
• D. ${\log }_e\left(5x\right)$

Answer 1

The correct option is B ${\log }_{e}x$

$arc\tanh \left(x\right)=\frac{1}{2}\ln \frac{(1+x)}{(1-x)}$
Therefore, substituting x by $\frac{x^2-1}{x^2+1}$
Therefore
$=\frac{1}{2}\ln \frac{(1+\frac{x^2-1}{x^2+1})}{(1-\frac{x^2-1}{x^2+1})}$
$=\frac{1}{2}\ln \frac{(x^2+1+x^2-1)}{(x^2+1-(x^2-1\left)\right)}$
$=\frac{1}{2}\ln \frac{\left(2x^2\right)}{\left(2\right)}$
$=\frac{1}{2}\ln x^2$
$=\ln \left(x\right)$