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

The inverse of f(x)=1+ln((1-x)/(1+x)) is -(e^{x-1}-1)/(e^{-1+x)+1}

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

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

\frac{x ^{3} + 8}{7}

f (x) = \frac{3}{7} x - 12

f (x) = \frac{3 ^{x - 2}}{5}

x - 3 + 9

f (x) = 2 x^{2} - 12 x + 5

What is the inverse of f(x)=1+ln((1-x)/(1+x)) ?
The inverse of f(x)=1+ln((1-x)/(1+x)) is -(e^{x-1}-1)/(e^{-1+x)+1}