Question:

$\textcolor{green}{\textbf{Key Idea}}$ Split the integrand: \[ \frac{2x(1+\sin x)}{1+\cos^2x} =\frac{2x}{1+\cos^2x} +\frac{2x\sin x}{1+\cos^2x}. \] The first term is odd and integrates to \(0\) over \([-\pi,\pi]\). The second term is even. Then use the symmetry \[ f(\pi-x)=f(x) \] for \[ f(x)=\frac{\sin x}{1+\cos^2x} \] on \([0,\pi]\). $\textcolor{blue}{\textbf{Solution}}$ Let \[ I=\int_{-\pi}^{\pi} \frac{2x(1+\sin x)}{1+\cos^2 x}\,dx. \] Split it as \[ I=\int_{-\pi}^{\pi}\frac{2x}{1+\cos^2x}\,dx +\int_{-\pi}^{\pi}\frac{2x\sin x}{1+\cos^2x}\,dx. \] The first integrand is odd, so its integral is \(0\). The second integrand is even, so \[ I=4\int_0^\pi \frac{x\sin x}{1+\cos^2x}\,dx. \] Let \[ J=\int_0^\pi \frac{x\sin x}{1+\cos^2x}\,dx. \] Since \[ \frac{\sin(\pi-x)}{1+\cos^2(\pi-x)} =\frac{\sin x}{1+\cos^2x}, \] we use the symmetry formula: \[ J=\frac{\pi}{2}\int_0^\pi \frac{\sin x}{1+\cos^2x}\,dx. \] Now put \[ u=\cos x,\qquad du=-\sin x\,dx. \] Then \[ \int_0^\pi \frac{\sin x}{1+\cos^2x}\,dx =\int_{-1}^{1}\frac{du}{1+u^2} =\left[\tan^{-1}u\right]_{-1}^{1} =\frac{\pi}{2}. \] Thus \[ J=\frac{\pi}{2}\cdot \frac{\pi}{2} =\frac{\pi^2}{4}. \] Therefore \[ I=4J=\pi^2. \] \[ \boxed{\pi^2} \] $\textcolor{red}{\textbf{Key Trap}}$ Do not declare the whole integrand odd because of the factor \(x\). The factor \(1+\sin x\) breaks that. Only \[ \frac{2x}{1+\cos^2x} \] is odd. The term \[ \frac{2x\sin x}{1+\cos^2x} \] is even and contributes the full answer.