Finding the value of this sum.

We want to find the value of this series $\sum_{n=1}^{\infty} \frac{n}{2^n(n+1)}$ According to Wolfram, tha value is $2 - ln(4)$

What I did :

$\frac{1}{1-x}$ = $\sum_{n=0}^{\infty} x^n$. Differentiation:

$\frac{1}{(1-x)^2}$ = $\sum_{n=1}^{\infty} nx^{n-1}$

$\frac{x}{(1-x)^2}$ = $\sum_{n=1}^{\infty} nx^{n}$ . Integration:

$\frac{1}{(1-x)}+ ln(\vert x-1 \vert) $ = $\sum_{n=1}^{\infty} \frac{nx^{n+1}}{n+1}$

$\frac{1}{x(1-x)}+ \frac{ln(\vert x-1 \vert)}{x} $ = $\sum_{n=1}^{\infty} \frac{nx^{n}}{n+1}$

Doing $x = \frac{1}{2}$, we have the desired series. But the function at $x= \frac{1}{2}$ is $2,61…$, not $2 - ln(4)$

What I am doing wrong?