Finding the inverse of $f(x) = x^3 + x$

In general, the inverse of a invertible closed-form function need not itself have a closed form. Simple examples of this include $x \mapsto x^5 + x$ and a suitable restriction of $x \mapsto x^x$.

Since the function $f$ is a cubic polynomial function, we can find an explicit inverse of the given function, $f(x) := x^3 + x$, by writing $y = x^3 + x$ using Cardano's Formula or the equivalent to solve for $x$ and hence inverting the given function. The (somewhat unpleasant) result is: $$\color{#bf0000}{\boxed{f^{-1}(y) = \frac{1}{6} \left(108 y+12 \sqrt{81 y^2+12}\right)^{1/3} - \frac{2}{\left(108 y+12 \sqrt{81 y^2+12}\right)^{1/3}}}} .$$

See Jack D'Aurizio's nice answer for a more compact way to write this function, using $\sinh$ and $\operatorname{arsinh}$.

Like all quadratic polynomial functions, some cubic polynomial functions are not one-to-one and hence do not admit a (global) inverse. A cubic polynomial $a x^3 + b x^2 + c x + d$, $a \neq 0$, is invertible iff its derivative does not have two real roots (which would correspond to local extrema of the cubic polynomial), and this is the case iff the discriminant of the derivative is nonpositive, or equivalently, iff $$b^2 \leq 3 a c .$$

I would like to write an addendum to Travis' answer.

By Lagrange's inversion formula, the solution $x$ of $x^3-x+z=0$ can be written as: $$ x = \sum_{k\geq 0}\binom{3k}{k}\frac{z^{2k+1}}{2k+1} $$ under the assumption $|z|< \frac{2}{\sqrt{27}} $ that makes such a series convergent.

It follows that the inverse function $g(y)$ of $f(x)=x^3+x$ can be represented through: $$ g(y) = \sum_{k\geq 0}\binom{3k}{k}\frac{(-1)^k y^{2k+1}}{2k+1} \tag{1}$$ for any $y\in\left(-\frac{2}{\sqrt{27}},\frac{2}{\sqrt{27}}\right)$.

The connection with the cubic formula (and the classical method for solving third-degree equations through the sine/cosine triplication formula) lies here:

$$ g(y) = \frac{2}{\sqrt{3}}\,\sinh\left[\frac{1}{3}\,\text{arcsinh}\left(\frac{3\sqrt{3}y}{2}\right)\right]\tag{2} $$

and the interesting fact is that both $(1)$ and $(2)$ give more compact formulas compared to Cardano's one involving $\sqrt[3]{\cdot}$.