How to evaluate the following indefinite integral? $\int e^{e^x}\mathrm dx$ I stumbled across this question: what's the value of the following integral? $$\int e^{e^x}\mathrm dx.$$ Furthermore I was required to demonstrate. On wolfram I got the result $\operatorname{Ei}(e^x)$ but I don't know what is exponential integral neither what to do with it.
Thank you
 A: There are functions that are integrable - i.e. an integral exists and is well defined, but we cannot express the integral in terms of elementary functions. Yours is one example of this.
The crux is that we can come up with infinitely many different functions, but we only have a finite number of types elementary functions. As such, there are many functions where an integral exists and is well behaved, but doesn't its integral take the form of any function that we've given a name to.
One solution to this is to create new function names - the Ei function is an example of just that, and is defined to be $$\text{Ei}(x)=-\int_{-x}^\infty\frac{e^{-t}}tdt$$
So that $$\frac d{dx}\mathrm{Ei}(x) = \frac{e^{x}}x$$We can then play with this function to define other integrals. For example, $$\begin{align}\frac d{dx}\text{Ei}(e^x) &= \frac{e^{e^x}}{e^x}\frac{d}{dx}e^x \qquad\text{(by the chain rule)}\\&=e^{e^x}\end{align}$$which is the example you were looking for. 
A: As Mathmo's answer makes clear, this isn't the kind of function for which you can find an elementary anti-derivative. But it is quite possible to approximate it: the Taylor series has a well-known form, and can be integrated term-by-term: $$e^{e^x-1}=\sum\limits_{n=0}^\infty \dfrac{B_n}{n!}x^n
\implies \int e^{e^x}\,dx = \sum_{n=1}^\infty \left(e\frac{B_{n-1}}{n!}\right)x^{n}+C$$ where $B_n$ is the $n$th Bell Number. These integers are well understood; you can look up a table of them, or use the method in the Wikipedia link to find as many as you need. In principle, then, one may use this formula to approximate any definite integral of this function arbitrarily well. (In practice, of course, it could take a lot of terms and one may be well-advised to accelerate the convergence of the series somehow.)
