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?

Another approach, if you already know the expansion $\ln(1+x) = x-\frac{x^2}{2}+\frac{x^3}{3} - \frac{x^4}{4}+\cdots$, is to literally multiply the series by $(1+x)$ and combine terms:
\begin{eqnarray*}
(1+x)\ln(1+x) & = & 1\cdot(x-\tfrac{x^2}{2}+\tfrac{x^3}{3}\cdots)+x\cdot(x-\tfrac{x^2}{2}+\tfrac{x^3}{3}\cdots)\\
& = & x + (\tfrac11-\tfrac12)x^2 -(\tfrac12-\tfrac13)x^3+(\tfrac13-\tfrac14)x^4-\cdots \\
& = & x + \tfrac1{1\cdot2}x^2-\tfrac1{2\cdot 3}x^3 + \tfrac1{3\cdot 4}x^4\cdots \\
& = & x + \tfrac{x^2}2 - \tfrac{x^3}{6} +\tfrac{x^4}{12} - \cdots
\end{eqnarray*}
