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

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

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inverse of x/(1+x^2)

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

\frac{7 + 4 x}{6 - 5 x}

f (x) = x + 4

\frac{x ^{2} + x + 2}{x - 1}

\frac{5 x}{x + 8} - 8

y = \frac{2}{5} x - 5

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