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

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

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inverse of (s+3)/((s^2-9)(s+1))

inverse

\frac{s + 3}{( s ^{2} - 9 ) ( s + 1 )}

\frac{s + s ( 4 s + 1 )}{s}, \frac{s - s ( 4 s + 1 )}{s}

Solution steps

y = \frac{s + 3}{( s ^{2} - 9 ) ( s + 1 )}

Replace

s

with

y

s = \frac{y + 3}{( y ^{2} - 9 ) ( y + 1 )}

Solve for

y, s = \frac{y + 3}{( y ^{2} - 9 ) ( y + 1 )}

\frac{s + s ( 4 s + 1 )}{s}, \frac{s - s ( 4 s + 1 )}{s}

f (x) = - 2 x^{2} + 8 x - 7, x > 2

lo g_{2} (11 x) + 4

f (x) = 4 cos (π x + \frac{π}{2})

7^{t - 2}

f (x) = \frac{4}{4 - x}

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