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: $$\ln(1+x) = \int \left(\frac{1}{1+x}\right)dx$$
$$\ln(1+x) = \sum_{k=0}^{\infty} \int (-x)^k dx$$

You got the general expansion about $x=a$. Here we are intended to take $a=0$. That is, we are finding the Maclaurin series of $\ln(1+x)$.
That will simplify your expression considerably. Note also that $\frac{(n-1)!}{n!}=\frac{1}{n}$.
The approach in the suggested solution also works. We note that
$$\frac{1}{1+t}=1-t+t^2-t^3+\cdots\tag{1}$$
if $|t|\lt 1$ (infinite geometric series). Then we note that
$$\ln(1+x)=\int_0^x \frac{1}{1+t}\,dt.$$
Then we integrate the right-hand side of (1) term by term.  We get
$$\ln(1+x) = x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots,$$
precisely the same thing as what one gets by putting $a=0$ in your expression.
