What is the inverse of f(x)=ln((x^2+1)/(x^2-1)) ?

The inverse of f(x)=ln((x^2+1)/(x^2-1)) is sqrt((1+e^x)/(e^x-1)),-sqrt((1+e^x)/(e^x-1))

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inverse of f(x)=ln((x^2+1)/(x^2-1))

inverse

f (x) = ln (\frac{x ^{2} + 1}{x ^{2} - 1})

\frac{1 + e ^{x}}{e ^{x} - 1}, - \frac{1 + e ^{x}}{e ^{x} - 1}

Solution steps

y = ln (\frac{x ^{2} + 1}{x ^{2} - 1})

Replace

x

with

y

x = ln (\frac{y ^{2} + 1}{y ^{2} - 1})

Solve for

y, x = ln (\frac{y ^{2} + 1}{y ^{2} - 1})

\frac{1 + e ^{x}}{e ^{x} - 1}, - \frac{1 + e ^{x}}{e ^{x} - 1}

\frac{2 x}{x ^{2} - 1}

f (x) = \frac{e ^{x}}{x ^{8}}

f (x) = x^{4} - 72 x^{2} + 9

\frac{4}{x ^{2} + x - 2}

f (x) = \frac{1 + x ^{2}}{4}

What is the inverse of f(x)=ln((x^2+1)/(x^2-1)) ?
The inverse of f(x)=ln((x^2+1)/(x^2-1)) is sqrt((1+e^x)/(e^x-1)),-sqrt((1+e^x)/(e^x-1))