taylor series of ln(1+x)?

Compute the taylor series of $ln(1+x)$

I’ve first computed derivatives (upto 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 answer is:
As $ln(1+x) = \int (\frac{1}{1+x})dx$
$ln(1+x) = \Sigma_{k=0}^{\infty} \int (-x)^k dx$

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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
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
precisely the same thing as what one gets by putting $a=0$ in your expression.

Note that
$$\frac{1}{1+x}=\sum_{n \ge 0} (-1)^nx^n$$
Integrating both sides gives you
&=\sum_{n \ge 0}\frac{(-1)^nx^{n+1}}{n+1}\\
&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\\
We deduce that
&=\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}\\
Edit: To derive a closed for for the geometric series, let
To prove in the other direction, use the binomial theorem or simply compute the series about $0$ manually.