taylor series of $\ln(1+x)$? Compute the taylor series of $\ln(1+x)$
I've first computed derivatives (up to  the 4th) of ln(1+x)
$f^{'}(x)$ = $\frac{1}{1+x}$ 
$f^{''}(x) = \frac{-1}{(1+x)^2}$
$f^{'''}(x) = \frac{2}{(1+x)^3}$
$f^{''''}(x) = \frac{-6}{(1+x)^4}$
Therefore the series:
$\ln(1+x) = f(a) + \frac{1}{1+a}\frac{x-a}{1!} - \frac{1}{(1+a)^2}\frac{(x-a)^2}{2!} +  \frac{2}{(1+a)^3}\frac{(x-a)^3}{3!} -  \frac{6}{(1+a)^4}\frac{(x-a)^4}{4!} + ...$
But this doesn't seem to be correct. Can anyone please explain why this doesn't work?
The supposed correct answers are: $$\ln(1+x) = \int \left(\frac{1}{1+x}\right)dx$$
$$\ln(1+x) = \sum_{k=0}^{\infty} \int (-x)^k dx$$

Note that
$$\frac{1}{1+x}=\sum_{n \ge 0} (-1)^nx^n$$
Integrating both sides gives you
\begin{align}
\ln(1+x)
&=\sum_{n \ge 0}\frac{(-1)^nx^{n+1}}{n+1}\\
&=x-\frac{x^2}{2}+\frac{x^3}{3}-...
\end{align}
Alternatively,
\begin{align}
&f^{(1)}(x)=(1+x)^{-1} &\implies \ f^{(1)}(0)=1\\
&f^{(2)}(x)=-(1+x)^{-2} &\implies f^{(2)}(0)=-1\\
&f^{(3)}(x)=2(1+x)^{-3} &\implies \ f^{(3)}(0)=2\\
&f^{(4)}(x)=-6(1+x)^{-4} &\implies \ f^{(4)}(0)=-6\\
\end{align}
We deduce that
\begin{align}
f^{(n)}(0)=(-1)^{n-1}(n-1)!
\end{align}
Hence,
\begin{align}
\ln(1+x)
&=\sum_{n \ge 1}\frac{f^{(n)}(0)}{n!}x^n\\
&=\sum_{n \ge 1}\frac{(-1)^{n-1}(n-1)!}{n!}x^n\\
&=\sum_{n \ge 1}\frac{(-1)^{n-1}}{n}x^n\\
&=\sum_{n \ge 0}\frac{(-1)^{n}}{n+1}x^{n+1}\\
&=x-\frac{x^2}{2}+\frac{x^3}{3}-...
\end{align}
Edit: To derive a closed for for the geometric series, let
\begin{align}
S&=1-x+x^2-x^3+...\\
xS&=x-x^2+x^3-x^4...\\
S+xS&=1\\
S&=\frac{1}{1+x}\\
\end{align}
To prove in the other direction, use the binomial theorem or simply compute the series about $0$ manually.
