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One of my student asked me to estimate this indefinite integral \begin{gather*} \int \frac{1}{e^x+x}dx. \end{gather*} I tried many methods, but I can not find the primitive of it. Then I use Maple and Mathematica to help me. As a result, the two softwares can not give me the result. Thus I am not sure if the primitive of this integral can be expressed as elementary functions. But I could not give the right reason. How can I judge if the primitive of this indefinite integral can be expressed as elementary functions? Can anybody help me?

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Is this a homework problem? – Mhenni Benghorbal Dec 4 '12 at 9:43
Downvoted for lack of attempt at answer and lack of research effort. – Epictetus Dec 4 '12 at 9:50
It is not totally a homework problem, but part of a textbook example. One of my student proposed this problem when she study that example, and actually I can not understand why she come across this problem? And I do not know how to using Liouville's Theorem to this problem! – nuage Dec 4 '12 at 9:51
For lots of info on elementary integrals, see… – Gerry Myerson Dec 4 '12 at 11:37
up vote 1 down vote accepted

It is very unlikely that the function $$f(x):=\int_0^x{dt\over e^t+t}$$ can be expressed in elementary terms. To prove this, however, would be quite difficult, for the following reasons: In the first place one needs a deep algebraic theory to deal with such questions, and even having this theory at your disposal it is not obvious how to handle the particular example considered here. This is the reason why 99% of the students are just told that you cannot express the really important function $\Phi(x):={1\over\sqrt{2\pi}}\int_{-\infty}^x e^{-t^2/2} dt$ in elementary terms, and never see a proof of this fact.

Now in your question you talk about "estimating" the function $x\mapsto f(x)$. This is another matter, and if you are really interested in an "estimate" you should indicate for which range of $x$ you want such an estimate to be useful.

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Thank you, Christian Blatter! Your answer is very helpful! – nuage Dec 6 '12 at 5:51

Case $1$: $xe^{-x}\leq1$

Then $\int\dfrac{1}{e^x+x}dx$



$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+1}n!x^ke^{-(n+1)x}}{(n+1)^{n-k+1}k!}+C$ (can be obtained from

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