Solution to $\int\frac{1}{x+xe^{ax+b}}dx$
I need to solve $\int\frac{1}{x+xe^{ax+b}}dx$ as part of a larger optimization problem, where the $a$ and $b$ parameters will be optimized in a subsequent step. Wolfram alpha returns the following Puiseux expansion centered at $x=0$:
$$\frac{\log{x}}{e^b+1} - \frac{ae^bx}{(e^b+1)^2} + O(x^2)$$
My questions
-
Is it possible to shift this expansion to some other value, say $x=c$, with a simple substitution, or would I need to re-calculate the series centered around a different point?
-
Are there better approaches to getting an approximate solution, perhaps sidestepping the series expansion? For example, would Simpson's rule, Gaussian quadrature or some other numerical technique give me a solution in terms of $a$ and $b$ that I can plug into subsequent calculations easier?
I am hoping someone with good intuition regarding numerical integration can point me in the right direction. The integrand seems to me to be a reasonably well behaved function, other than the singularity at $0$, so I feel like I should be able to get an approximation that will not make my downstream optimization task too difficult.
As an aside, I only care about positive values of $x$, namely in the $(0,200]$ interval.
It is just a bit more complex. Expanding as series around $x=c\neq0$, the integrand write $$\frac{1}{c (1+e^{a c+b})}\Bigg[1+\sum_{n=1}^p \alpha_n (x-c)^n\Bigg]+O\left((x-c)^{p+1}\right)$$ and the very first coefficients are $$\alpha_1=-\frac{(a c+1) e^{a c+b}+1}{c \left(e^{a c+b}+1\right)}$$ $$\alpha_2=\frac{\left((a c+1) e^{a c+b}+1\right)^2}{c^2\left( e^{a c+b}+1\right)^2}-\frac{a (a c+2) e^{a c+b}}{2 c\left( e^{a c+b}+1\right)}$$ The next ones are too long to type but you can get them using Wolfram Alpha