How to show that $\int_{0}^{\pi}(1+2x)\cdot{\sin^3(x)\over 1+\cos^2(x)}\mathrm dx=(\pi+1)(\pi-2)?$ How do we show that?

$$\int_{0}^{\pi}(1+2x)\cdot{\sin^3(x)\over 1+\cos^2(x)}\mathrm dx=(\pi+1)(\pi-2)\tag1$$

$(1)$ it a bit difficult to start with
$$\int_{0}^{\pi}(1+2x)\cdot{\sin(x)[1-\sin^2(x)]\over 1+\cos^2(x)}\mathrm dx\tag2$$
Setting $u=\cos(x)$ 
$du=-\sin(x)dx$
$$\int_{-1}^{1}(1+2x)\cdot{(u^2)\over 1+u^2}\mathrm du\tag3$$
$$\int_{-1}^{1}(1+2\arccos(u))\cdot{(u^2)\over 1+u^2}\mathrm du\tag4$$
$du=\sec^2(v)dv$
$$\int_{-\pi/4}^{\pi/4}(1+2\arccos(\tan(v)))\tan^2(v)\mathrm dv\tag5$$
$$\int_{-\pi/4}^{\pi/4}\tan^2(v)+2\tan^2(v)\arccos(\tan(v))\mathrm dv=I_1+I_2\tag6$$
$$I_1=\int_{-\pi/4}^{\pi/4}\tan^2(v)\mathrm dv=2-{\pi\over2}\tag7$$
As for $I_2$ I am sure how to do it.

I am not sure this is the easiest (but it is a viable approach). Note that the antiderivative of $v'=\sin^3 x/(1+\cos^2 x)$ is given by $v=\cos x -2 \arctan \cos x$. Thus, we can integrate the problem by parts and obtain
$$I = (\cos x -2 \arctan \cos x)(1+2x)\Big|_{x=0}^\pi -2 \underbrace{\int_0^\pi(\cos x -2 \arctan \cos x)dx}_{=I_2}.$$
Now you can observe that $v=\cos x -2 \arctan \cos x$ is antisymmetric around $x=\pi/2$ (because $\cos$ is antisymmetric around $x=\pi/2$ and $\arctan$ is antisymmetric around $x=0$). Thus, the remaining integral vanishes $(I_2=0)$ and we obtain
$$I = (\cos x -2 \arctan \cos x)(1+2x)\Big|_{x=0}^\pi= (\pi+1)(\pi-2).$$
