What is the inverse of 1/(s^2+4s+5) ?

The inverse of 1/(s^2+4s+5) is (-2s+sqrt(s(-s+1)))/s ,-(2s+sqrt(s(-s+1)))/s

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inverse of 1/(s^2+4s+5)

inverse

\frac{1}{s ^{2} + 4 s + 5}

\frac{- 2 s + s ( - s + 1 )}{s}, - \frac{2 s + s ( - s + 1 )}{s}

Solution steps

y = \frac{1}{s ^{2} + 4 s + 5}

Replace

s

with

y

s = \frac{1}{y ^{2} + 4 y + 5}

Solve for

y, s = \frac{1}{y ^{2} + 4 y + 5}

\frac{- 2 s + s ( - s + 1 )}{s}, - \frac{2 s + s ( - s + 1 )}{s}

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

\frac{( 3 s + 1 )}{( s ^{2} + s - 12 )}

y = \frac{2 x}{x - 1}

f (x) = (3)

f (x) = \frac{8 x}{3 x - 2}

What is the inverse of 1/(s^2+4s+5) ?
The inverse of 1/(s^2+4s+5) is (-2s+sqrt(s(-s+1)))/s ,-(2s+sqrt(s(-s+1)))/s