Official solution
\[\begin{aligned}
& I=\int_\limits{-\pi}^\pi \frac{2 y(1+\sin y)}{1+\cos ^2 y} d y \\
& =\int_\limits0^\pi\left(\frac{2 y(1+\sin y)}{1+\cos ^2 y}+\frac{-2 y(1-\sin y)}{1+\cos ^2 y}\right) d y \\
& =\int_\limits0^\pi\left(\frac{2 y+2 y \sin y-2 y+2 y \sin y}{1+\cos ^2 y}\right) d y
\end{aligned}\]
\[I=4 \int_0^\pi\left(\frac{y \sin }{1+\cos ^2 y}\right) d y \quad \text{... (1)}\]
\[\begin{aligned}
& I=4 \int_\limits0^\pi\left(\frac{(\pi-y) \sin (\pi-)}{1+\cos ^2(\pi-y)}\right) d y \\
& I=4\left\lfloor\int_\limits0\left(\frac{\pi \sin y}{1+\cos ^2 y}\right) d y-\int_\limits0^\pi \frac{y \sin y}{1+\cos ^2 y} d y\right\rfloor \quad \text{... (2)}
\end{aligned}\]
Adding equation (1) and (2)
\[\begin{aligned}
& 2 I=4 \int_\limits0^\pi\left(\frac{\pi \sin y}{1+\cos ^2 y}\right) d y \\
& I=2 \pi \int_\limits0^\pi \frac{\sin y}{1+\cos ^2 y} d y \\
& =2 \pi \times \frac{\pi}{2} \\
& =\pi^2
\end{aligned}\]