1. We need to evaluate $I=\int_{-\pi}^{\pi} \frac{2y(1+\sin y)}{1+\cos^2 y}\,dy.$

2. Split the integrand: $\frac{2y(1+\sin y)}{1+\cos^2 y}=\frac{2y}{1+\cos^2 y}+\frac{2y\sin y}{1+\cos^2 y}.$ So, $I=\int_{-\pi}^{\pi} \frac{2y}{1+\cos^2 y}\,dy+\int_{-\pi}^{\pi} \frac{2y\sin y}{1+\cos^2 y}\,dy.$

3. Use parity (odd/even properties):

• $y$ is odd.
• $\cos^2 y$ is even, so $1+\cos^2 y$ is even.
• Hence $\dfrac{2y}{1+\cos^2 y}$ is odd.

Therefore, $\int_{-\pi}^{\pi} \frac{2y}{1+\cos^2 y}\,dy=0.$

4. Now consider $\frac{2y\sin y}{1+\cos^2 y}.$ Here:

• $y$ is odd,
• $\sin y$ is odd,
• product $y\sin y$ is even,
• denominator $1+\cos^2 y$ is even. So the whole function is even.

Thus, $I=\int_{-\pi}^{\pi} \frac{2y\sin y}{1+\cos^2 y}\,dy=2\int_0^{\pi} \frac{2y\sin y}{1+\cos^2 y}\,dy=4\int_0^{\pi} \frac{y\sin y}{1+\cos^2 y}\,dy.$

5. Let $J=\int_0^{\pi} \frac{y\sin y}{1+\cos^2 y}\,dy.$ Use integration by parts:

• Take $u=y \Rightarrow du=dy$,
• Take $dv=\dfrac{\sin y}{1+\cos^2 y}dy$.

Now compute $v$: Let $t=\cos y$, then $dt=-\sin y\,dy$. So, $\int \frac{\sin y}{1+\cos^2 y}dy=-\int \frac{dt}{1+t^2}=-\tan^{-1}(t)=-\tan^{-1}(\cos y).$ Hence, $v=-\tan^{-1}(\cos y).$

Therefore, $J=\left[-y\tan^{-1}(\cos y)\right]_0^{\pi}+\int_0^{\pi} \tan^{-1}(\cos y)\,dy.$

6. Evaluate the boundary term:

• At $y=\pi$, $\cos\pi=-1$, so $\tan^{-1}(-1)=-\frac{\pi}{4}$. Thus, $-\pi\tan^{-1}(\cos\pi)=-\pi\left(-\frac{\pi}{4}\right)=\frac{\pi^2}{4}.$
• At $y=0$, the term is $0$.

So, $J=\frac{\pi^2}{4}+\int_0^{\pi} \tan^{-1}(\cos y)\,dy.$

7. Now evaluate $K=\int_0^{\pi} \tan^{-1}(\cos y)\,dy.$ Use the substitution $y\mapsto \pi-y$:

Since $\tan^{-1}(-x)=-\tan^{-1}(x)$, $K=-\int_0^{\pi} \tan^{-1}(\cos y)\,dy=-K.$ Hence, $K=0.$

Therefore, $J=\frac{\pi^2}{4}.$

8. Finally, $I=4J=4\cdot \frac{\pi^2}{4}=\pi^2.$

9. So the correct option is $\boxed{\pi^2}$ which is option B.