What is the inverse of (s+3)/((s+1)(s+4)) ?

The inverse of (s+3)/((s+1)(s+4)) is (-5s+1+sqrt(9s^2+2s+1))/(2s),(-5s+1-sqrt(9s^2+2s+1))/(2s)

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inverse of (s+3)/((s+1)(s+4))

inverse

\frac{s + 3}{( s + 1 ) ( s + 4 )}

\frac{- 5 s + 1 + 9 s ^{2} + 2 s + 1}{2 s}, \frac{- 5 s + 1 - 9 s ^{2} + 2 s + 1}{2 s}

Solution steps

y = \frac{s + 3}{( s + 1 ) ( s + 4 )}

Replace

s

with

y

s = \frac{y + 3}{( y + 1 ) ( y + 4 )}

Solve for

y, s = \frac{y + 3}{( y + 1 ) ( y + 4 )}

\frac{- 5 s + 1 + 9 s ^{2} + 2 s + 1}{2 s}, \frac{- 5 s + 1 - 9 s ^{2} + 2 s + 1}{2 s}

f (x) = \frac{( x + 2 )}{3 x + 1}

\frac{2 s + 3}{s ^{2} + 9}

h (t) = - 16 t^{2} + 30 t + 5

f (x) = 1.25 (38 - x)

y = 9 x

What is the inverse of (s+3)/((s+1)(s+4)) ?
The inverse of (s+3)/((s+1)(s+4)) is (-5s+1+sqrt(9s^2+2s+1))/(2s),(-5s+1-sqrt(9s^2+2s+1))/(2s)