Question

lf $|x|<1$, then $\frac{1}{2}\log \left(\frac{1+x}{1-x}\right)=$

• A. $\tanh x$
• B. ${\sinh }^{-1}x$
• C. ${\cosh }^{-1}x$
• D. ${\tanh }^{-1}x$

Answer 1

The correct option is D ${\tanh }^{-1}x$

Formula $tan{h}^{-1}x=\frac{1}{2}ln\left(\frac{1+x}{1-x}\right)$
explanation -2
$y=\frac{1}{2}ln\left(\frac{1+x}{1-x}\right)$
$2y=ln\left(\frac{1+x}{1-x}\right)$
$e^{2y}=\frac{1+x}{1-x}$
$x=\frac{e^{2y}-1}{e^{2y}+1}=\frac{e^y-e^{-y}}{e^y+e^{-y}}=\frac{sinhy}{coshy}=tanhy$
$\therefore tan{h}^{-1}x=y$