How can I find the $n$'th derivative of $f(x)=x^2 \ln(x+1)$ 

How can I find the $n$'th derivative of $f(x)=x^2 \ln(x+1)$


I have tried this 
$$f(x)=x^2\ln(x+1)$$
$$ f'(x) = 2x\ln(x+1) + \frac{x^2}{x+1}$$
$$ f''(x)=2\ln(x+1) + \frac{4x}{x+1} - \frac{x^2}{(x+1)^2}$$
However I do not see any pattern :( 
 A: Taking one more derivative results in
$$
f^{(3)}(x) = \frac2{x+1} + \frac{(x+1)4-4x(1)}{(x+1)^2} - \frac{(x+1)^22x-x^22(x+1)}{(x+1)^4} = \frac{2 \left(x^2+3 x+3\right)}{(x+1)^3}.
$$
However, that's not the most convenient form for taking further derivatives: using partial fractions gives us
$$
f^{(3)}(x) = \frac{2}{x+1}+\frac{2}{(x+1)^2}+\frac{2}{(x+1)^3}.
$$
And this should now be easy to take as many derivatives of as you want.
A: Taylor series gives $$f(x)=x^2\ln(1+x)=x^2\sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^n}{n}$$
If you want to get the 2016th derivative at x=0, it will be $$\frac{d^{2016}}{dx}((x^2)(-1)^{2014}\frac{x^{2014}}{2014})=\frac{2016!}{2014}$$
A: Let $u=x^2$ and $v=\ln (1+x)$ then,
$$u'=2x,u''=2,u'''=0$$
And
$v^{(n)}=\frac{(-1)^{n-1}(n-1)!}{(1+x)^n}$ for $n \geq 1$.
Let's compute $(uv)^{(3)}=\sum_{k=0}^{3} {3 \choose k} u^{(k)}v^{(n-k)}={3 \choose 0} x^2\frac{(-1)^{n-1}(n-1)!}{(1+x)^n}+{3 \choose 1}2x\frac{(-1)^{n-2}(n-2)!}{(1+x)^{n-1}}+{3 \choose 2}2\frac{(-1)^{n-3}(n-3)!}{(1+x)^{n-2}}+{3 \choose 3}0$
With $n=3$ by the general product rule
Now it's easy to see that for $n \geq 3$ we have 
$$(uv)^{(n)}={n \choose 0} x^2\frac{(-1)^{n-1}(n-1)!}{(1+x)^n}+{n \choose 1}2x\frac{(-1)^{n-2}(n-2)!}{(1+x)^{n-1}}+{n \choose 2}2\frac{(-1)^{n-3}(n-3)!}{(1+x)^{n-2}}$$
As if the third derivative of $u$ is $0$ ,and the fourth, the fifth, the sixth, etc. 
This simplifies to,
$$x^2\frac{(-1)^{n-1}(n-1)!}{(1+x)^n}+2x\frac{(-1)^{n-2}n!}{(n-1)(1+x)^{n-1}}+\frac{(-1)^{n-3}n!}{(n-2)(1+x)^{n-2}}$$
