Find the closed form for this series I found this interesting series from the , it is from an old math books. It is as followed:
$\dfrac{1}{2}-\dfrac{x^2}{6}+\dfrac{x^4}{12}-\dfrac{x^6}{20}+\dfrac{x^8}{30}-...$
I notice that one can rewrite this series as followed:
$\dfrac{1}{2}-\dfrac{x^2}{2\cdot 3}+\dfrac{x^4}{3\cdot 4}-\dfrac{x^6}{4\cdot 5}+\dfrac{x^8}{5\cdot 6}-...$
So the general formula for this series is 
$$\sum_{n=0}^{\infty} \dfrac{(-1)^{n}x^{2n}}{(n+1)(n+2)}$$
Is there a closed form for this series?

$$\sum_{n=0}^{\infty} \dfrac{(-1)^{n}x^{2n}}{(n+1)(n+2)}=\sum_{n=0}^\infty(-1)^nx^{2n}\int_0^1\int_0^1 y^nz^{n+1}dydz$$
$$=\int_0^1\int_0^1z\sum_{n=0}^\infty(-x^2yz)^ndydz$$
$$=\int_0^1\int_0^1\frac{z}{1+x^2yz}dydz$$
$$=\int_0^1z\left(\int_0^1\frac{dy}{1+x^2yz}\right)dz$$
$$=\int_0^1z\left(\frac{\ln(1+x^2z}{x^2z}\right)dz$$
$$=\frac1{x^2}\int_0^1\ln(1+x^2z)dz$$
$$=\frac1{x^2}\cdot\frac{(1+x^2)\ln(1+x^2)-x^2}{x^2}$$
