Answer
$f\left( x \right) = x\ln \left( {1 + \dfrac{x}{2}} \right) = \mathop \sum \limits_{n = 2}^\infty \dfrac{{{{\left( { - 1} \right)}^n}}}{{\left( {n - 1} \right){2^{n - 1}}}}{x^n}$
Work Step by Step
We compute the derivatives of $f\left( x \right) = x\ln \left( {1 + \dfrac{x}{2}} \right)$ and list them in the table below:
$\begin{array}{*{20}{c}}
n&{\dfrac{{{f^{\left( n \right)}}\left( x \right)}}{{n!}}}&{\dfrac{{{f^{\left( n \right)}}\left( 0 \right)}}{{n!}}}\\
0&{x\log \left( {\dfrac{x}{2} + 1} \right)}&0\\
1&{\dfrac{{x + x\ln \left( {\dfrac{x}{2} + 1} \right) + 2\ln \left( {\dfrac{x}{2} + 1} \right)}}{{x + 2}}}&0\\
2&{\dfrac{{x + 4}}{{2{{\left( {x + 2} \right)}^2}}}}&{\dfrac{1}{2}}\\
3&{\dfrac{{ - x - 6}}{{6{{\left( {x + 2} \right)}^3}}}}&{ - \dfrac{1}{8}}\\
4&{\dfrac{{x + 8}}{{12{{\left( {x + 2} \right)}^4}}}}&{\dfrac{1}{{24}}}\\
5&{\dfrac{{ - x - 10}}{{20{{\left( {x + 2} \right)}^5}}}}&{ - \dfrac{1}{{64}}}
\end{array}$
By Theorem 1 of Section 11.7, the Taylor series centered at $c=0$ is given by
$T\left( x \right) = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{f^{\left( n \right)}}\left( 0 \right)}}{{n!}}{x^n}$
Using the table above, we get
$T\left( x \right) = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{f^{\left( n \right)}}\left( 0 \right)}}{{n!}}{x^n} = \dfrac{1}{2}{x^2} - \dfrac{1}{8}{x^3} + \dfrac{1}{{24}}{x^4} - \dfrac{1}{{64}}{x^5} + \cdot\cdot\cdot$
We can write the series as
$T\left( x \right) = \mathop \sum \limits_{n = 2}^\infty \dfrac{{{{\left( { - 1} \right)}^n}}}{{\left( {n - 1} \right){2^{n - 1}}}}{x^n}$
Thus, the Taylor series of $f\left( x \right) = x\ln \left( {1 + \dfrac{x}{2}} \right)$ centered at $c=0$ is
$f\left( x \right) = x\ln \left( {1 + \dfrac{x}{2}} \right) = \mathop \sum \limits_{n = 2}^\infty \dfrac{{{{\left( { - 1} \right)}^n}}}{{\left( {n - 1} \right){2^{n - 1}}}}{x^n}$
