Express $y=\sinh ^{-1}x$ as a natural logarithm.

Solution Since $y=\sinh ^{-1}x$, where $x=\sinh y,$ we have $$\begin{eqnarray*} x &=&\dfrac{e^{y}-e^{-y}}{2} \\ 2xe^{y} &=&( e^{y}) ^{2}-1\qquad {\color{#0066A7}{\hbox{Multiply both sides by }2e^{y}.}} \\ ( e^{y}) ^{2}-2xe^{y}-1 &=&0 \end{eqnarray*}$$

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This is a quadratic equation in $e^{y}$. Use the quadratic formula and solve for $e^{y}$. $$\begin{eqnarray*} e^{y} &=&\dfrac{2x\pm \ \sqrt{4x^{2}+4}}{2}\qquad \\ e^{y} &=&x\pm \ \sqrt{x^{2}+1}\qquad\ {\color{#0066A7}{\hbox{ Simplify.}}} \end{eqnarray*}$$

Since $e^{y}>0$ and $x<\ \sqrt{x^{2}+1}$ for all $x$, the minus sign on the right side is not possible. As a result, $e^{y}=x+\ \sqrt{x^{2}+1}$ so that $$y=\sinh ^{-1}x=\ln ( x+\ \sqrt{x^{2}+1})$$