Hint Taylor Series $x^{2}\ln(x)$ about $a=1$ Find the taylor series of $g(x):= x^{2}\ln(x)$ about $a=1$
Idea: I would have thought we should find the Taylor series of $\ln(x)$ about $a=1$, which is 
$\sum^{\infty}_{k=1}\frac{(-1)^{k+1}}{k!}(x-1)^{k}$ and then I would just multiply that with $x^{2}$. So, we'd get:
$\sum^{\infty}_{k=1}\frac{(-1)^{k+1}}{k!}(x-1)^{k}x^2$
This cannot be simplified can it? Surely there must be a better without having to calculate the $n$-th derivative of $x^2\ln(x)$

Your answer  is  no completely correct. As you would like the expansion around $a=1$, you should also expand $x^2$ around this point. In fact, you can easily show that
$$x^2 = 1 +2 (x-1) + (x-1)^2\;.$$
As a result, we have that
$$x^2 \ln x = \Bigl[1 +2 (x-1) + (x-1)^2 \Bigr] \sum_{k=1}^\infty \frac{(-1)^{k+1}}k (x-1)^k\;.$$
Or, after expanding
$$x^2 \ln x = \sum_{k=1}^\infty \frac{(-1)^{k+1}}k (x-1)^k + 2 \sum_{k=1}^\infty \frac{(-1)^{k+1}}k (x-1)^{k+1} + \sum_{k=1}^\infty \frac{(-1)^{k+1}}k (x-1)^{k+2}\\
= \sum_{k=1}^\infty \frac{(-1)^{k+1}}k (x-1)^k + 2 \sum_{k=2}^\infty \frac{(-1)^{k}}{k-1} (x-1)^{k} + \sum_{k=3}^\infty \frac{(-1)^{k+1}}{k-2} (x-1)^{k}\\
= (x-1) + \frac32 (x-1)^2 +\sum_{k=3}^\infty  (-1)^{k+1}\left[\frac{1}{k} - \frac{2}{k-1} + \frac{1}{k-2} \right] (x-1)^k.$$ 
