Does $\int \frac{e^x(x-1)}{1+xe^x}dx$ have any closed form? It's simple that $$\int \frac{e^x(x+1)}{1+xe^x}dx=\ln(1+xe^x)+C,$$
But what if $$\int \frac{e^x(x-1)}{1+xe^x}dx?$$
 A: This will not be a closed form, but still works. There may be a complicated formula with Meijer G related functions for a closed form.
You can see what happens with a simple numerator conjugate. This answer will be inspired from similar questions. It uses geometric series and gamma functions. This representation works only for $x<W(1)=Ω=.5671…$. This is called the W-lambert function representation of the Omega constant. We can integrate term by term as the sum converges. Notice the gamma function and exponential integral function argument in the integrand. This is an integration source:
$$\int \frac{e^x(x-1)}{xe^x+1}dx= \int e^x(x-1) \frac{1}{1-\left(-xe^x\right)}dx =\int e^x(x-1)\sum_{y=0}^\infty \left(-xe^x\right)^ydx= \sum_{y=0}^\infty (-1)^y\int x^{y+1} e^{xy+x} -x^y e^{xy+x} dx= C+\sum_{y=0}^\infty -\frac{e^{-x y} (xe^x )^y (x (y + 1))^{-y} ((y + 1) Γ(y + 1, -x (y + 1)) + (-1)^y Γ(y + 2, -x (y + 1)))}{(y + 1)^2}\mathop =^{0\le x<Ω, x\in \Bbb R} \sum_{y=0}^\infty \left[-(1 + y)^{-1 - y} Γ(1 + y, -x (1 + y)) - (1 + y)^{-2 - y} Γ(2 + y, -x (1 + y))\right] = C+ \sum_{y=0}^\infty (-1)^y x^{1 + y} \left(\frac{e^{x (1 + y)}}{1 + y} + 2\text E_{-y}( -x (1 + y))\right) $$
If there are ways to simplify using any academically recognized special function, please feel free to simplify. Please correct me and give me feedback!
