evaluate the summation : $\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(n+1)(n+2x+3)}$ 
Find
$$S=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(n+1)(n+2x+3)}$$
for $x≥0$.

At first, I use a partial fraction
$$S=\displaystyle\sum_{n=0}^{\infty}\left(\frac{(-1)^{n}}{2(x+1)(n+1)}-\frac{(-1)^{n}}{2(x+1)(n+2x+3)}\right)
=\frac{1}{2(x+1)}(I-J),$$
where
$$I=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n+1}
=\ln 2$$
and
$$J=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n+3+2x}$$
I think to use:
$$\ln (1+y)=\sum_{n=0}^{\infty}\frac{(-1)^{n}y^{n+1}}{n+1}$$
Is my work correct? How to complete this work ?

$$S(x)=\sum_{n=0}^{\infty} \frac{(-1)^n}{(n+1)(n+2x+3)}=\sum_{n=0}^{\infty}
\frac{(-1)^n}{n+1} \int _{0}^{1} t^{n+2x+2} dt=\int_{0}^{1} t^{2x+1} \ln(1+t) dt ~~~(1).$$ $$\Rightarrow S(x)=\left .\ln(1+t) \frac{t^{2x+2}}{2x+2}\right |_{0}^{1}-\frac{1}{2x+2}\int_{0}^{1} \frac{t^{2x+2}dt}{1+t}=\frac{\ln 2}{2x+2} - \frac{\beta(2x+3)}
{2x+2}~~~~(2).$$
Here we have used 3.222 of Garadshteyn and Ryzhik where $$\int_{0}^{1}\frac{x^{{\mu}-1}}{1+x} dx= \beta(\mu).$$
Next $$\beta(x)=\frac{1}{2}[\psi(\frac{x+1}{2})-\psi(\frac{x}{2})]~~~~(3)$$
and $\psi(x)$ is Digamma function definded as $\psi(x)=\frac{\Gamma'(x)}
{\Gamma(x)}= -C -\sum \left( \frac{1}{x+k}-\frac{1}{k+1} \right),$ $C$ is Eular constant equa to =$-\psi(1)=0.5772....$
Using (3), we find that $$ S(0)=0.25, S(1)=0.1458..=7/48, S(2)=0.1027=37/360, S(3)=0.7931=533/6720.$$ 
