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 :( 

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.
