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inverse of (e^x - e^-x)/2
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Set u = e^x and create a quadratic in u. Solve and you'll find the function (arcsinh) you want.
put \[y=\frac{e^x+e^{-x}}{2}\] and solve for x via \[2y=e^x-\frac{1}{e^x}=\frac{e^{2x}-1}{e^x}\] \[2ye^x=e^{2x}-1\] \[e^{2x}-2ye^x-1=0\] and solve this quadratic equation in \[e^x\] as jamesj wrote
I.e., if y = (1/2)(e^x - e^-x) then y = (1/2)(u - 1/u) Rearrange to obtain an equation in u. Then substitute back in e^x.
Right, crossed messages ... what sat73 just wrote as well.
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