What is the inverse of D/(Dx)x^3+x+1 ?

The inverse of D/(Dx)x^3+x+1 is (-1+sqrt(4x-3))/2 ,(-1-sqrt(4x-3))/2

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inverse of D/(Dx)x^3+x+1

inverse

\frac{D}{D x} x^{3} + x + 1

\frac{- 1 + 4 x - 3}{2}, \frac{- 1 - 4 x - 3}{2}

Solution steps

y = \frac{D}{D x} x^{3} + x + 1

Replace

x

with

y

x = \frac{D}{Dy} y^{3} + y + 1

Solve for

y, x = \frac{D}{Dy} y^{3} + y + 1

\frac{- 1 + 4 x - 3}{2}, \frac{- 1 - 4 x - 3}{2}

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

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

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

f (x) = \frac{3 x}{7 x - 2}

h (x) = \frac{- 2}{3 + x}

What is the inverse of D/(Dx)x^3+x+1 ?
The inverse of D/(Dx)x^3+x+1 is (-1+sqrt(4x-3))/2 ,(-1-sqrt(4x-3))/2