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

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

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

inverse

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

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

Solution steps

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

Replace

s

with

y

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

Solve for

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

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

f (x) = \frac{1}{2} (x^{2} - 4)

f (x) = 8 x^{3} - 7

f (x) = - 3 + ln (x)

f (x) = 3 x + 4

f (x) = 20 - 20 e^{- l n (10) \cdot t}

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