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I`m trying to evaluate this integral $$\int\limits_0^1\frac{(1-x)e^x}{x+e^x}\,dx.$$ Would you please give me any idea?

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Wolfram can't find a closed form, but it approximates it to $0.417559$. What are you hoping for exactly? –  1015 Jun 3 '13 at 13:46
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$$\int_{0}^{1}e^xd\ln{\left(\dfrac{x}{e^x}+1\right)}$$ –  math110 Jun 3 '13 at 13:49
    
@julien: I wish I could evaluate it exactly and represent it in natural form. –  Takasima Senko Jun 3 '13 at 13:52
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@math110 That's a very nice observation, but what can we do with that? –  1015 Jun 3 '13 at 13:59
    
@julien Thank you, follow I have no any idea. –  math110 Jun 3 '13 at 14:23
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1 Answer

Let's expand the integrand as follows:

$$\frac{(1-x)e^x}{x+e^x}=\frac{(1-x)}{1+xe^{-x}}=(1-x)(1-xe^{-x}+x^2e^{-2x}-...)=$$

$$=(1-x)\sum_{i=0}^{\infty}(-1)^ix^ie^{-ix}=$$

$$=\sum_{i=0}^{\infty}(-1)^ix^ie^{-ix}-\sum_{i=0}^{\infty}(-1)^ix^{i+1}e^{-ix}$$

Next, we need the following result:

$$I(m,k)=\int_{0}^{1}x^me^{-kx}dx=$$

$$=\frac{m!}{k^{m+1}}-e^{-k}\sum_{j=0}^{m}j!\binom{m}{j}\frac{1}{k^{j+1}};\;m\geqslant n$$

The integral can be evaluated by integration by parts.

Using this result, the original integral can be expressed in terms of $I(m,k):$

$$\int\limits_0^1\frac{(1-x)e^x}{x+e^x}\,dx=\sum_{k=0}^{\infty}(-1)^k\left [ I(k,k)-I(k+1,k) \right ]$$

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