$$(2n+1)^3-(2n+1)=(2n+1)((2n+1)^2-1)$$
$$=(2n+1)(4n^2+4n)=4n(n+1)(2n+1)$$
$$ \Rightarrow \sum_{n=1}^ \infty \frac{4(-1)^{n+1}}{(2n+1)^3-(2n+1)}=\sum_{n=1}^ \infty \frac{4(-1)^{n+1}}{4n(n+1)(2n+1)}$$
$$=\sum_{n=1}^ \infty\frac{(-1)^{n+1}}{n(n+1)(2n+1)}=\sum_{n=1}^ \infty \frac{(-1)^{n+1}}{(2n+1)}( \frac{1}{n}- \frac{1}{n+1})$$
$$=\sum_{n=1}^ \infty \frac{2(-1)^{n+1}}{(2n+1)}.\frac{1}{2n}-\sum_{n=1}^ \infty \frac{2(-1)^{n+1}}{(2n+1)}.\frac{1}{2(n+1)}$$
$$=2\sum_{n=1}^ \infty(-1)^{n+1}( \frac{1}{2n}- \frac{1}{2n+1})-2\sum_{n=1}^ \infty(-1)^{n+1}(\frac{1}{2n+1} -\frac{1}{2n+2})$$
$$=\sum_{n=1}^ \infty\frac{(-1)^{n+1}}{n}-4\sum_{n=1}^ \infty\frac{(-1)^{n+1}}{2n+1}+\sum_{n=1}^ \infty\frac{(-1)^{n+1}}{n+1}$$
$$=Ln2-4(1-\frac{\pi}{4})-Ln2+1$$
$$=\pi-3$$
$$=\pi-[\pi]$$
$\Box$
