Solve and give correct answer for this 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)
${\displaystyle f\prime \left(x\right)=\frac{1}{1+x}}$
${\displaystyle f\prime \prime \left(x\right)=\frac{-1}{{(1+x)}^2}}$
${\displaystyle f\prime \prime \prime \left(x\right)=\frac{2}{{(1+x)}^3}}$
${\displaystyle f\prime \prime \prime \prime \left(x\right)=\frac{-6}{{(1+x)}^4}}$
Therefore the series:
${\displaystyle \ln (1+x)=f\left(a\right)+\frac{1}{1+a}\frac{x-a}{1!}-\frac{1}{{(1+a)}^2}\frac{{(x-a)}^2}{2!}+\frac{2}{{(1+a)}^3}\frac{{(x-a)}^3}{3!}-\frac{6}{{(1+a)}^4}\frac{{(x-a)}^4}{4!}+\dots }$
But this doesn't seem to be correct. Can anyone please explain why this doesn't work?
The supposed correct answer is:
${\displaystyle As\ln (1+x)=\int \left(\frac{1}{1+x}\right)dx}$
${\displaystyle \ln (1+x)=\sum _{k=0}^{\mathrm{\infty }}\int {(-x)}^{k}dx}$

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 ${\displaystyle \frac{(n-1)!}{n!}=\frac{1}{n}.}$
The approach in the suggested solution also works. We note that
${\displaystyle \frac{1}{1+t}=1-t+t^2-t^3+\cdots \left(1\right)}$
if ${\displaystyle \left|t\right|<1}$ (infinite geometric series). Then we note that
${\displaystyle \ln (1+x)={\int }_0^x\frac{1}{1+t}dt.}$
Then we integrate the right-hand side of (1) term by term. We get
${\displaystyle \ln (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots ,}$
precisely the same thing as what one gets by putting a=0 in your expression.

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