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

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

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

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

f (x) = 3 x^{4}

- 3 sin (2 π x + 2)

y = - tan (x - \frac{π}{3})

5 x - y = 9

f (x) = \frac{1}{x} + 2 ln (x + 3)

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