What is the inverse of (s+1)/(s(s-3)) ?

The inverse of (s+1)/(s(s-3)) is (3s+1+sqrt(9s^2+10s+1))/(2s),(3s+1-sqrt(9s^2+10s+1))/(2s)

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inverse of (s+1)/(s(s-3))

inverse

\frac{s + 1}{s ( s - 3 )}

\frac{3 s + 1 + 9 s ^{2} + 10 s + 1}{2 s}, \frac{3 s + 1 - 9 s ^{2} + 10 s + 1}{2 s}

Solution steps

y = \frac{s + 1}{s ( s - 3 )}

Replace

s

with

y

s = \frac{y + 1}{y ( y - 3 )}

Solve for

y, s = \frac{y + 1}{y ( y - 3 )}

\frac{3 s + 1 + 9 s ^{2} + 10 s + 1}{2 s}, \frac{3 s + 1 - 9 s ^{2} + 10 s + 1}{2 s}

Y (x) = (x + 1)^{2}

- 5 x^{2} + x, x \geq 1

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

f (x) = \frac{7 x + 1}{- 4 x + 9}

f (x) = 0.008 x^{3}

What is the inverse of (s+1)/(s(s-3)) ?
The inverse of (s+1)/(s(s-3)) is (3s+1+sqrt(9s^2+10s+1))/(2s),(3s+1-sqrt(9s^2+10s+1))/(2s)