<s>[INST] Taylor series of $(1+x)\ln(1+x)$ in $x=0$ How to determine the Taylor series of $(1+x)\ln(1+x)$ in $x=0$?
My idea is finding the second derivative of the expression, which is $\frac{1}{1+x}$. 
The Taylor series of this expression is $1-x+x^2-x^3$ and so on. If I integrate then this series I get $\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{12}-\frac{x^5}{20}$ and so on. But the solution is  $x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{12}-\frac{x^5}{20}$. My question is, how does the $x$ at the beginning of the series appear?
 [/INST] You might want to use the fact that 
$$
    \frac{\mathrm{d}^2}{\mathrm{d}x^2} \left( (1+x) \ln(1+x) \right) = 
   \frac{\mathrm{d}}{\mathrm{d}x} \left( \ln(1+x) + 1  \right) = \frac{1}{1+x} = \sum_{n=0}^\infty (-1)^n x^n
$$
Integrating term-wise twice, and using initial conditions of 0 for the function, and 1 for the first derivative:
$$
   (1+x) \ln(1+x) = x + \sum_{n=0}^\infty \frac{(-1)^n x^{n+2}}{(n+1)(n+2)}
$$
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