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?
 A: 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:
$$
   (1+x) \ln(1+x) = x + \sum_{n=0}^\infty \frac{(-1)^n x^{n+2}}{(n+1)(n+2)}
$$
A: Each time you integrate (or, more properly, antidifferentiate), you will introduce an arbitrary constant. You need to account for those.
A: $\newcommand{\+}{^{\dagger}}%
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 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
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 \newcommand{\dd}{{\rm d}}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\fermi}{\,{\rm f}}%
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
 \newcommand{\half}{{1 \over 2}}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
 \newcommand{\ol}[1]{\overline{#1}}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}%
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
 \newcommand{\sech}{\,{\rm sech}}%
 \newcommand{\sgn}{\,{\rm sgn}}%
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
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 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
Since
$\ds{\ln\pars{1 + x} = \sum_{n = 1}^{\infty}\pars{-1}^{n + 1}\,{x^{n} \over n}}$

\begin{align}
\color{#0000ff}{\large\pars{1 + x}\ln\pars{1 + x}}
&=\sum_{n = 1}^{\infty}\pars{-1}^{n + 1}\,{x^{n} \over n}
+
\sum_{n = 1}^{\infty}\pars{-1}^{n + 1}\,{x^{n + 1} \over n}
\\[3mm]&=\sum_{n = 1}^{\infty}\pars{-1}^{n + 1}\,{x^{n} \over n}
+
\sum_{n = 2}^{\infty}\pars{-1}^{n}\,{x^{n} \over n - 1}
\\[3mm]&=x + \sum_{n = 2}^{\infty}\bracks{{\pars{-1}^{n + 1} \over n} +
{\pars{-1}^{n} \over n - 1}}x^{n}
=\color{#0000ff}{\large x + \sum_{n = 2}^{\infty}
{\pars{-1}^{n} \over \pars{n - 1}n}\,x^{n}}
\\[3mm]&=
\color{#0000ff}{\large x + {x^{2} \over 2} - {x^{3} \over 2\cdot 3}
+  {x^{4} \over 3\cdot 4} - {x^{5} \over 4\cdot 5} + {x^{6} \over 5\cdot 6} - \cdots}
\end{align}

A: 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*}
