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It's simple that $$\int \frac{e^x(x+1)}{1+xe^x}dx=\ln(1+xe^x)+C,$$ But what if $$\int \frac{e^x(x-1)}{1+xe^x}dx?$$

mengdie1982
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    I don't think that the latter integral has a closed form (yet). Note that, if it would have a closed form, then so would be the sum of these two integrals which is $2\int\frac{xe^x}{1+xe^x}dx$ and this one doesn't have a closed form either. – RMWGNE96 May 09 '19 at 13:15
  • Yes. We can study $\int\frac{1}{1+xe^x}dx$ – mengdie1982 May 09 '19 at 13:37
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    Wolfram Dev Platform doesn't produce anything useful, so I'm going to say "no". – Adrian Keister May 09 '19 at 14:59
  • @AdrianKeister Hi, I used wolframalpha.com to solve the sextic $56y^6-336y^5+378y^4+819y^3-2079y^2+1512y-351 = 0$ which does not give closed form. But actually it can be solved in closed form since it can be factorized into a product of two cubics. Why can wolframalpha not do such a easy task? – River Li Aug 20 '21 at 23:34
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    It seems like Wolfram Alpha only evaluates integrals in closed form which have the functions from the Wolfram functions site. This means asking it to evaluate the integral of the integral in the question only gives a simple series expansion not in Σ notation. – Тyma Gaidash Aug 20 '21 at 23:36
  • @RiverLi WolframAlpha or Wolfram Dev Platform are limited, though they can do quite a lot. Even WolframAlpha can't solve the general quintic (proven impossible), nor can it find an elementary antiderivative of $e^{-x^2}.$ – Adrian Keister Aug 21 '21 at 01:25
  • @AdrianKeister Thanks. I understand they can't solve general quintic since impossible. Some quintic equations can be solved by radicals (use Galois group solvable command to know it). For these equations, it is a simple task to give closed form just as giving closed form for cubic. It is simpler to give the decomposition of a product of two cubics (if solvable). However, they do not do it. – River Li Aug 21 '21 at 01:41
  • @TymaGaidash Sorry, I mean if the roots of sextic equations (not general case) can be expressed by radicals, why wolfram alpha do not give them (just as easy as giving roots of cubic). For example, it is an easy job to give closed form of $56y^6-336y^5+378y^4+819y^3-2079y^2+1512y-351 = 0$. – River Li Aug 21 '21 at 01:47
  • @TymaGaidash No, I mean the sextic equations which can be factorized into the product of two cubics, rather than general cases. – River Li Aug 21 '21 at 01:51
  • I see. In that case just factor. – Тyma Gaidash Aug 21 '21 at 01:54
  • @TymaGaidash I tried to factor it in wolframalpha.com, without desired output. One can do it by hand: $$56y^6-336y^5+378y^4+819y^3-2079y^2+1512y-351 = 56, \left( {y}^{3}-3,{y}^{2}+ \left( -{\frac {9}{8}}-3/8,\sqrt {105 } \right) y+{\frac {63}{16}}+{\frac {51}{112}},\sqrt {105} \right) \left( {y}^{3}-3,{y}^{2}+ \left( -{\frac {9}{8}}+3/8,\sqrt {105} \right) y+{\frac {63}{16}}-{\frac {51}{112}},\sqrt {105} \right) $$ – River Li Aug 21 '21 at 02:17

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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!

Тyma Gaidash
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