Prove that $\sum\limits_{n=0}^\infty\binom{2n}n\frac n{4^n(n+1)^2}=\ln(16)-2$ 
Prove that
  $$\sum\limits_{n=0}^\infty\binom{2n}n\frac n{4^n(n+1)^2}=\ln(16)-2$$

According to Wolfram the above holds. Could someone show me the steps for this?

HINT
First note that
$$\sum_{n\geq0}\binom{2n}n\frac n{4^n(n+1)^2}=\sum_{n\geq0}\binom{2n}n\frac{(n+1)-1}{4^n(n+1)^2}=\sum_{n\geq0}\binom{2n}n\frac1{4^n(n+1)}-\sum_{n\geq0}\binom{2n}n\frac1{4^n(n+1)^2}$$
Now recall that
$$\sum_{n\geq0}\binom{2n}n\frac{x^n}{4^n}=\frac1{\sqrt{1-x}},~~~\text{for }|x|<1$$
The task boils down to an exercise in integration. Can you take it from here?

For the interested reader I will add the complete solution now. As given, note that for the first sum we have the following
$$\sum_{n\geq0}\binom{2n}n\frac1{4^n(n+1)}=\sum_{n\geq0}\binom{2n}n\frac1{4^n}\int_0^1x^n{\rm d}x=\int_0^1\frac{{\rm d}x}{\sqrt{1-x}}=[-2\sqrt{1-x}]_0^1=2$$
This also gives us the following identity
$$\sum_{n\geq0}\binom{2n}n\frac{x^n}{4^n(n+1)}=\int\frac{{\rm d}x}{\sqrt{1-x}}=-2\sqrt{1-x}+c$$
Take $x=0$ to see that $c=2$. Dividing by $x$ and integrating again over $[0;1]$ gives
\begin{align*}
\sum_{n\geq0}\binom{2n}n\frac1{4^n(n+1)^2}&=\int_0^1\frac{2-2\sqrt{1-x}}x{\rm d}x\\
&=2\int_0^1\frac{1-\sqrt{1-x}}x{\rm d}x\\
&=2\int_0^1\frac{1-\sqrt x}{1-x}{\rm d}x\\
&=2\int_0^1\frac1{1+\sqrt x}{\rm d}x\\
&=2\sum_{n\geq0}(-1)^n\int_0^1x^{\frac n2}{\rm d}x\\
&=2\sum_{n\geq0}\frac{(-1)^n}{\frac n2+1}\\
&=4\sum_{n\geq0}\frac{(-1)^n}{n+2}\\
&=4\left[\sum_{n\geq1}\frac{(-1)^n}n+1\right]\\
&=4-4\log2
\end{align*}
The result follows.

$$\therefore~\sum_{n\geq0}\binom{2n}n\frac n{4^n(n+1)^2}~=~4\log2-2$$

