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

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

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

inverse

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

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

Solution steps

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

Replace

s

with

y

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

Solve for

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

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

f (x) = 3 x^{2} - 3 x - 18

cos (- 0.15106)

f (x) = \frac{8}{3 + x ^{2}}

f (x) = - ln (x + 2) + 1

f (x) = 4 + 4 x - 8

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