What is the inverse of 1/(4s^2+1) ?

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

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inverse of 1/(4s^2+1)

inverse

\frac{1}{4 s ^{2} + 1}

\frac{s 1 - s}{2 s}, - \frac{s 1 - s}{2 s}

Solution steps

y = \frac{1}{4 s ^{2} + 1}

Replace

s

with

y

s = \frac{1}{4 y ^{2} + 1}

Solve for

y, s = \frac{1}{4 y ^{2} + 1}

\frac{s 1 - s}{2 s}, - \frac{s 1 - s}{2 s}

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