What is the inverse of (s+1)/(s^2-4s+13) ?

The inverse of (s+1)/(s^2-4s+13) is (4s+1+sqrt(-36s^2+12s+1))/(2s),(4s+1-sqrt(-36s^2+12s+1))/(2s)

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inverse of (s+1)/(s^2-4s+13)

inverse

\frac{s + 1}{s ^{2} - 4 s + 13}

\frac{4 s + 1 + - 36 s ^{2} + 12 s + 1}{2 s}, \frac{4 s + 1 - - 36 s ^{2} + 12 s + 1}{2 s}

Solution steps

y = \frac{s + 1}{s ^{2} - 4 s + 13}

Replace

s

with

y

s = \frac{y + 1}{y ^{2} - 4 y + 13}

Solve for

y, s = \frac{y + 1}{y ^{2} - 4 y + 13}

\frac{4 s + 1 + - 36 s ^{2} + 12 s + 1}{2 s}, \frac{4 s + 1 - - 36 s ^{2} + 12 s + 1}{2 s}

y = \frac{2 x}{3 x - 5}

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

f (x) = ln (7 - 4 x)

y = 0.2 sin (x)

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

What is the inverse of (s+1)/(s^2-4s+13) ?
The inverse of (s+1)/(s^2-4s+13) is (4s+1+sqrt(-36s^2+12s+1))/(2s),(4s+1-sqrt(-36s^2+12s+1))/(2s)