indefinite integral computation $dx/(e^{-x}-x)$ I am trying to carry out the following indefinite integral:
$$\int \frac{1}{e^{-q} - q} \, dq$$
Mathematica is not helping me, and i think it is not solvable by substitution method.
any idea on how to solve it?
 A: According to Wolfram: $\displaystyle{\frac{1}{e^{-x}-x}=1+2x+\frac{7x^2}{2}+\frac{37x^3}{6}+\cdots}$ 
therefore the integral can be written like that : 
$$\displaystyle{\int \frac{1}{e^{-x}-x}\, dx=\int \left(1+2x+\frac{7x^2}{2}+\frac{37x^3}{6}+\cdots \right )\, dx=x+x^2+\frac{7x^3}{6}+\frac{37x^4}{24}+\cdots}$$
A: According to http://calculus-geometry.hubpages.com/hub/List-of-Functions-You-Cannot-Integrate-No-Antiderivatives, the function
$$f: x \mapsto \frac{1}{e^x + x}$$ does not have an antiderivative expressible in terms of elementary functions, i.e. as algebraic combinations and compositions of polynomials, $\ln$, and $\exp$. Your integrand is simply $f(-x)$.
However, many functions of this kind have indefinite integrals, for instance $$\int_{0}^{\infty} \frac{\mathrm{d}x}{e^x + 1} = \ln(2).$$ Mathematica tells us there is no such expression for your integrand, however.
A: It may not be possible to get a nice clean answer. To me, it looks like your best bet is to use a Taylor series.
To refresh your memory, the Taylor series of a function centered at $x=a$ is: $$\sum _{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n,$$ where $f^{(n)}(a)$ is the $n^{\mathrm{th}}$ derivative of $f$ at $a$.
Since Taylor series are polynomials, it should be simple to come up with the indefinite integral of the Taylor series of your function.
A: Case $1$: $xe^x\leq1$
Then $\int\dfrac{1}{e^{-x}-x}dx$
$=\int\dfrac{e^x}{1-xe^x}dx$
$=\int\sum\limits_{n=0}^\infty x^ne^{(n+1)x}~dx$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n-k}n!x^ke^{(n+1)x}}{(n+1)^{n-k+1}k!}+C$ (can be obtained from http://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions)
