<s>[INST] Finding the Maclaurin series for $f(x)=x\ln(x+1)$ 
Find the Maclaurin series for the function
  $$f(x)=x\ln(x+1)$$

So finding the derivatives is the first step. How many derivatives I need to find is explicitly said so I'll just go till the $4^{th}$ derivative.
$$\begin{align}
f'(x)&=\frac{x}{x+1}+\ln(x+1) & f'(0) &=0 \\
f''(x)&=\frac{1}{(x+1)^2}+\frac{1}{x+1} & f''(0)&=2 \\
f^{(3)}(x)&=\frac{-x-3}{(x+1)^3} & f^{(3)}(0)&=0 \\
f^{(4)}(x)&=\frac{2x+8}{(x+1)^4} & f^{(4)}(0)&=8
\end{align}$$
So now plugging in for the Maclaurin form 
$$P_n(x)=0+x^2-\frac{1}{2}x^3+\frac{1}{3}x^4+....$$
Is this correct?
 [/INST] You may use
$$\frac1{1-t} = 1 + t + t^2 + \>...$$
to obtain,
$$x\ln(1+x) = -x\int_0^{-x} \frac{dt}{1-t}=-x\int_0^{-x}\left(1+t+t^2+\>...\right)dt
= x^2 - \frac{x^3}2 + \frac{x^4}3+\>...$$
</s>