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I am trying to evaluate integral:

$$\int \frac{x^2\log(x)}{x+1}dx$$

But I have some problems with it. If I use Wolfram Alpha like this I get a result, but I need evaluate it by hand. Which method should I use?

If we represent it as:

$$\int(x^2\log(x)\cdot d(\ln(x))dx$$

then it is unclear what to dot here, even we can represent as power, take $x^2$ in power of $\log(x)$, not hint yet. Please help me to evaluate it.

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1 Answer 1

up vote 6 down vote accepted

We have \begin{align*} \int\frac{x^2\log x}{x+1}dx&=\int x\log x\frac{x+1-1}{x+1}dx\\ &=\int x\log x-\int \frac{x+1-1}{x+1}\log xdx\\ &=\frac{x^2}2\log x-\int\frac{x^2}2\frac 1xdx-\int \log x+\int\frac{\log x}{x+1}dx\\ &=\frac{x^2}2\log x-\frac{x^2}4-x\log x+x+\log x\log(x+1)-\int\frac{\log(x+1)}xdx\\ &=\frac x4(4-x)+\log x\left(\frac{x^2}2-x+\log(x+1)\right)-\int\frac{\log(x+1)}xdx\\ &=\frac x4(4-x)+\log x\left(\frac{x^2}2-x+\log(x+1)\right)-\sum_{k=1}^{+\infty}\int \frac{(-1)^{k+1}x^k}{kx}dx\\ &=\frac x4(4-x)+\log x\left(\frac{x^2}2-x+\log(x+1)\right)+\sum_{k=1}^{+\infty}(-1)^k \frac{x^k}{k^2}\\ &=\frac x4(4-x)+\log x\left(\frac{x^2}2-x+\log(x+1)\right)+\operatorname{Li}_2(-x), \end{align*} which is the result given by Wolfram Alpha.

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The summation runs over $j$ while the summands are in terms of $k$, so, the infinite sum diverges. No? Or I guess it's a typo. [Didn't go through the answer, so apologies if wrong!] –  user21436 Jan 21 '12 at 11:48
    
@KannappanSampath You are right, it's a typo that I will correct readily. –  Davide Giraudo Jan 21 '12 at 12:39

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