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

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

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

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

5 x + 2 y = 14

f (x) = \frac{( x ^{3} )}{x ^{2} - 1}

\frac{x ^{2} - 25}{- 2 x ^{2} - 10 x}

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

\frac{5 - 2 x}{6 x + 3}

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