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

Start with the Taylor series of the derivative
$$\frac 1{1+x}=\frac 1{(1+a)+(x-a)}=\frac 1{1+a}\,\,\frac 1{ 1+\frac {x-a}{1+a}}$$ Let $t=\frac {x-a}{1+a}$
$$\frac 1{1+t}=\sum_{n=0}^\infty (-1)^n t^n$$ So
$$\frac 1{1+x}=\frac 1{1+a}\,\,\sum_{n=0}^\infty (-1)^n \left(\frac{x-a}{a+1}\right)^n$$
Now, integrate
$$\log(1+x)=\frac 1{1+a}\,\,\sum_{n=0}^\infty (-1)^n \int\left(\frac{x-a}{a+1}\right)^n \,dx=\frac 1{1+a}\,\,\sum_{n=0}^\infty (-1)^n \frac{1+a}{n+1} \left(\frac{x-a}{a+1}\right)^{n+1}+C$$ Simplify
$$\log(1+x)=\sum_{n=0}^\infty (-1)^n \frac{1}{n+1} \left(\frac{x-a}{a+1}\right)^{n+1}+C$$ and using $x=a$, then $C=\log(1+a)$
