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How to calculate $$\int_0^1 \frac{\ln(1+x)} x \, dx $$ I tried to use different methods of integral but i didn't find any result.

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    $\begingroup$ What about using the Taylor series for $\ln(1+x)$ and integrating term-by-term? $\endgroup$ – sharding4 Jun 15 '17 at 1:26
  • $\begingroup$ I didn't really understand you meant $\endgroup$ – user431856 Jun 15 '17 at 1:29
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We have

$$ \int \frac{\ln(1+x)}{x} dx = -\sum_{k=1}^\infty \frac{(-1)^k(x)^k}{k^2} $$

So your definite integral

\begin{align*} \int_0^1 \frac{\ln(1+x)}{x} dx &= \left[-\sum_{k=1}^\infty \frac{(-1)^k(x)^k}{k^2} \right]_0^1 \\ \left[-\sum_{k=1}^\infty \frac{(-1)^k}{k^2} - 0\right] &= - \left( - \frac{\pi^2}{12} \right) \\ &= \frac{\pi^2}{12} \end{align*}

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HINT

$$\int_{0}^{1} \frac{\ln(1+x)}{x} \ dx= \int_{0}^{1} \int_{0}^{1} \frac{1}{1+xy} \ dy \ dx,$$ which can be confirmed by integrating the right hand side with respect to $y.$ Then make the change of variables $x=u+v,y=u-v$ to the double integral and evaluate. You should get $\frac{\pi^2}{12}.$

Remark

This is the exact same method of proof that Tom Apostol used for showing $$\sum_{n=1}^{\infty} \frac{1}{n^2} =\frac{\pi^2}{6}.$$ See pages 10-11 of http://math.cmu.edu/~bwsulliv/MathGradTalkZeta2.pdf

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