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

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

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

inverse

\frac{s + 1}{s ^{3} + 1}

\frac{s + - 3 s ^{2} + 4 s}{2 s}, \frac{s - - 3 s ^{2} + 4 s}{2 s}

Solution steps

y = \frac{s + 1}{s ^{3} + 1}

Replace

s

with

y

s = \frac{y + 1}{y ^{3} + 1}

Solve for

y, s = \frac{y + 1}{y ^{3} + 1}

\frac{s + - 3 s ^{2} + 4 s}{2 s}, \frac{s - - 3 s ^{2} + 4 s}{2 s}

f (x) = 3 x^{3} - 10

y = (1 - x)^{2}

\frac{1}{( x ^{2} + 4 )}

y = 2 x + 7 + 5

f (x) = - 8 x - 8

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