I know that the derivative is $$\frac{1}{1+x}$$ and I use the geometric series to write it as
$$\sum\limits_{n=0}^{\infty} (-x)^n$$. Since it is a derivative of $$ ln(1+x)$$ I need to find an antiderivative and that is
$$\sum\limits_{n=0}^{\infty} \frac{(-x)^{n+1}}{n+1}$$
Is this right?
On
Almost. You integrate $x^n$ to get $\frac{x^{n+1}}{n+1}$, so that $(-x)^n = (-1)^n x^n$ becomes $(-1)^n\frac{x^{n+1}}{n+1}$, not $(-1)^{n+1}\frac{x^{n+1}}{n+1}$. (The minus sign does not "get integrated.")
The resulting series is $$ \ln(1+x) = \sum_{n=0}^\infty(-1)^n\frac{x^{n+1}}{n+1} = \sum_{n=1}^\infty(-1)^{n+1}\frac{x^{n}}{n} $$ for $\lvert x\rvert < 1$. A simple way to check at least the first two terms: $$ \ln(1+0) = 0 $$ and $$\lim_{x\to 0}\frac{\ln(1+x)}{x} = 1$$ so that the first two coefficients should be (and are) $0$ and $1$, respectively.
Try integrating both sides. Note that the expansion loses validity outside of the open interval $(-1, 1)$. Use the fact that $(-x)^{n} = (-1)^{n}x^n$. This will prevent the confusion with the sign.