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How can I integrate this? $\int_{0}^{1}\frac{\ln(x)}{x+1} dx $

I've seen this but I failed to apply it on my problem.

Could you give some hint?

EDIT : From hint of @H.H.Rugh, I've got $\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^2}$, since $\int_{0}^{1}x^{n}\ln(x)dx = (-1)\frac{1}{(n+1)^2}$. How can I proceed this calculation hereafter?

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user124697
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2 Answers2

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Hint: $\frac{1}{1+x}=1-x+x^2-x^3...$ Then integrate term by term using partial integration.

H. H. Rugh
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  • Thank you, I've got the answer $\sum_{n=1}^{\infty}(-1)^n \frac{1}{n^2}$. Could you give hints that how can I calculate this sum? – user124697 Oct 30 '16 at 08:57
  • You then use $(1-\frac{2}{2^2}) (1+\frac{1}{2^2}+\frac{1}{3^2} ...) = 1-\frac{1}{2^2}+\frac{1}{3^2}-...$ and you probably recognize the first series. – H. H. Rugh Oct 30 '16 at 09:01
  • Wow, thank you so much. I didn't know the last formula. – user124697 Oct 30 '16 at 09:05
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    Decompose it into the sum over even-indexed terms and the sum over odd-indexed terms. Both sums can be found pretty immediately if you know $\sum_{n=1}^{\infty} \frac{1}{n^2}$. – Vik78 Oct 30 '16 at 09:59
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You can derivate $Li_2(-x)+(\ln x)(\ln (x+1))+C$ where $\displaystyle \frac{d}{dz}Li_2(z)=-\frac{\ln(1-z)}{z}$ and you will see how to integrate.

Hint: $\displaystyle\enspace Li_s(z):=\sum\limits_{k=1}^\infty \frac{z^k}{k^s}$ and $\displaystyle\enspace Li_2(1)=\frac{\pi^2}{6}$ .

user90369
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