= Thm =
> $\sinh^{-1}x=\ln(x+\sqrt{x^2+1})$
$x\in\mathbb{R}$

= Pf =
 $y=\sinh^{-1}x$
 i.e.
 $\sinh y=x$
 $\frac{e^y-e^{-y}}2=x$
 $e^y-2x-e^{-y}=0$
$t=e^y$ 치환
 $t-2x-\frac1t=0$
 $t^2-2xt-1=0$
 $t=e^y=x\pm\sqrt{x^2+1}$
$t=e^y>0$ 이므로
 $e^y=x+\sqrt{x^2+1}$

Up: [[여러가지증명]]