Finding the Maclaurin series for $f(x)=x\ln(x+1)$

Question

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?

Answer 1

You may use

$$\frac1{1-t} = 1 + t + t^2 + \>...$$

to obtain,

$$x\ln(1+x) = -x\int_0^{-x} \frac{dt}{1-t}=-x\int_0^{-x}\left(1+t+t^2+\>...\right)dt = x^2 - \frac{x^3}2 + \frac{x^4}3+\>...$$

Answer 2

Note that $$\ln(1+x)=\sum_{k=1}^{\infty} \frac{(-1)^{k+1}x ^k}{k}$$ So the Mclaurin series for $x\ln(1+x)$ is $$ x\ln(1+x)= \sum_{k=1}^{n} \frac{(-1)^{k+1} x^{k+1}}{k}=x^2-\frac{x^3}{2}+\frac{x^4}{3}-\frac{x^5}{4}+....$$