Question:

<div class='solution'> <p>$\frac{2x(1+\sin x)}{1+\cos^2 x} = \frac{2x}{1+\cos^2 x}+\frac{2x\sin x}{1+\cos^2 x}$</p> <p><strong>Term 1:</strong> $\frac{2x}{1+\cos^2 x}$ is odd ($f(-x)=-f(x)$) → $\int_{-\pi}^\pi=0$.</p> <p><strong>Term 2:</strong> $\frac{2x\sin x}{1+\cos^2 x}$ — let $g(x)=\frac{x\sin x}{1+\cos^2 x}$. Check: $g(-x)=\frac{(-x)(-\sin x)}{1+\cos^2 x}=g(x)$ → even.</p> <p>$\int_{-\pi}^\pi g\,dx = 2\int_0^\pi g\,dx$.</p> <p>King on $\int_0^\pi$: $\int_0^\pi g\,dx=\int_0^\pi\frac{(\pi-x)\sin x}{1+\cos^2 x}dx$. Add: $2\int_0^\pi g\,dx=\pi\int_0^\pi\frac{\sin x}{1+\cos^2 x}dx$.</p> <p>Let $t=\cos x$: $=\pi\int_{-1}^1\frac{dt}{1+t^2}=\pi\cdot\frac{\pi}{2}=\frac{\pi^2}{2}$. So $\int_0^\pi g=\frac{\pi^2}{4}$.</p> <p>Total: $2\cdot\frac{\pi^2}{4}\cdot 2 = \pi^2$. $\boxed{\pi^2}$</p> </div>