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

The inverse of (s+1)/((s-1)^2+1) is (2s+1+sqrt(-4s^2+8s+1))/(2s),(2s+1-sqrt(-4s^2+8s+1))/(2s)

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

inverse

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

\frac{2 s + 1 + - 4 s ^{2} + 8 s + 1}{2 s}, \frac{2 s + 1 - - 4 s ^{2} + 8 s + 1}{2 s}

Solution steps

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

Replace

s

with

y

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

Solve for

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

\frac{2 s + 1 + - 4 s ^{2} + 8 s + 1}{2 s}, \frac{2 s + 1 - - 4 s ^{2} + 8 s + 1}{2 s}

y = \frac{1}{6} \cdot 1 0^{x - 1} + 5

f (x) = - x + 1

f (x) = x^{5} + 5

f (x) = \frac{66 x + 2}{22 x - 5}

x = \frac{- 1}{2 ( y - 1 ) 2 + 5}

What is the inverse of (s+1)/((s-1)^2+1) ?
The inverse of (s+1)/((s-1)^2+1) is (2s+1+sqrt(-4s^2+8s+1))/(2s),(2s+1-sqrt(-4s^2+8s+1))/(2s)