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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse of s/(s^2+1)

inverse

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

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

Solution steps

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

Replace

s

with

y

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

Solve for

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

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

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

arcsin (e^{x} - 1)

f (x) = 1 - cos (x)

f (x) = - 5 + \frac{1}{2} x

x^{8}

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