What is the inverse of f(x)=2ln(x^2+1) ?

The inverse of f(x)=2ln(x^2+1) is sqrt(e^{x/2)-1},-sqrt(e^{x/2)-1}

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inverse of f(x)=2ln(x^2+1)

inverse

f (x) = 2 ln (x^{2} + 1)

e^{\frac{x}{2}} - 1, - e^{\frac{x}{2}} - 1

Solution steps

y = 2 ln (x^{2} + 1)

Replace

x

with

y

x = 2 ln (y^{2} + 1)

Solve for

y, x = 2 ln (y^{2} + 1)

e^{\frac{x}{2}} - 1, - e^{\frac{x}{2}} - 1

f (x) = e^{- x}

x^{2} - 6 x

f (x) = \frac{2}{x + 4}

f (x) = \frac{3 x ^{3} - 1}{x ^{2} - 2 x}

f (x) = 4 x^{3} - 3 x

What is the inverse of f(x)=2ln(x^2+1) ?
The inverse of f(x)=2ln(x^2+1) is sqrt(e^{x/2)-1},-sqrt(e^{x/2)-1}