What is the inverse of F(s)=ln((s^2+1)/(s^2)) ?

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

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inverse of F(s)=ln((s^2+1)/(s^2))

inverse

F (s) = ln (\frac{s ^{2} + 1}{s ^{2}})

\frac{1}{e ^{s} - 1}, - \frac{1}{e ^{s} - 1}

Solution steps

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

Replace

s

with

y

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

Solve for

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

\frac{1}{e ^{s} - 1}, - \frac{1}{e ^{s} - 1}

f (x) = - 9 x - 8

- 3 (x + 4)^{2} - 1

(21) (34)

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

- \frac{4}{x - 1}

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