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

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

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inverse of (x+1)/(x^2+1)

inverse

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

\frac{1 + 1 - 4 x ( x - 1 )}{2 x}, \frac{1 - 1 - 4 x ( x - 1 )}{2 x}

Solution steps

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

Replace

x

with

y

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

Solve for

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

\frac{1 + 1 - 4 x ( x - 1 )}{2 x}, \frac{1 - 1 - 4 x ( x - 1 )}{2 x}

f (x) = 3^{2 x}

x^{2} + 2 x + 5

f (x) = - x^{5} - 1

f (x) = \frac{5 - 2 x}{3 x}

f (x) = \frac{10 x - 17}{16}

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