Question:

Step 1: Split the integral into two parts. We decompose the integrand by separating the numerator: $$I = \int_{-\pi}^{\pi} \frac{2x(1 + \sin x)}{1 + \cos^2 x}\, dx = \int_{-\pi}^{\pi} \frac{2x}{1+\cos^2 x}\, dx + \int_{-\pi}^{\pi} \frac{2x \sin x}{1+\cos^2 x}\, dx$$ Step 2: Analyze the parity of the first integral. The function $\frac{2x}{1+\cos^2 x}$ is a product of an odd function ($2x$) and an even function ($\frac{1}{1+\cos^2 x}$), making it odd overall. Since we are integrating an odd function over the symmetric interval $[-\pi, \pi]$: $$\int_{-\pi}^{\pi} \frac{2x}{1+\cos^2 x}\, dx = 0$$ Step 3: Analyze the parity of the second integral. The function $\frac{2x \sin x}{1+\cos^2 x}$ is a product of an odd function ($2x$) and an odd function ($\sin x$), making it even overall. For an even function over a symmetric interval: $$\int_{-\pi}^{\pi} \frac{2x \sin x}{1+\cos^2 x}\, dx = 2\int_0^{\pi} \frac{2x \sin x}{1+\cos^2 x}\, dx$$ Step 4: Apply the standard integral property for $\int_0^\pi x f(\sin x)\, dx$. For integrals of the form $\int_0^\pi x f(\sin x)\, dx$, we use the property: $$\int_0^\pi x f(\sin x)\, dx = \frac{\pi}{2}\int_0^\pi f(\sin x)\, dx$$ Applying this with $f(\sin x) = \frac{\sin x}{1+\cos^2 x}$: $$\int_0^{\pi} \frac{2x \sin x}{1+\cos^2 x}\, dx = 2 \cdot \frac{\pi}{2}\int_0^{\pi} \frac{\sin x}{1+\cos^2 x}\, dx = \pi\int_0^{\pi} \frac{\sin x}{1+\cos^2 x}\, dx$$ Step 5: Evaluate the remaining integral using substitution. Let $u = \cos x$, so $du = -\sin x\, dx$. When $x = 0$, $u = 1$; when $x = \pi$, $u = -1$: $$\pi\int_0^{\pi} \frac{\sin x}{1+\cos^2 x}\, dx = \pi\int_{1}^{-1} \frac{-du}{1+u^2} = \pi\int_{-1}^{1} \frac{du}{1+u^2}$$ Step 6: Evaluate the arctangent integral. $$\pi\int_{-1}^{1} \frac{du}{1+u^2} = \pi\left[\tan^{-1}(u)\right]_{-1}^{1} = \pi\left(\tan^{-1}(1) - \tan^{-1}(-1)\right)$$ $$= \pi\left(\frac{\pi}{4} - \left(-\frac{\pi}{4}\right)\right) = \pi \cdot \frac{\pi}{2} = \frac{\pi^2}{2}$$ Step 7: Combine results to find the final answer. From Step 3, the total integral is: $$I = 2\int_0^{\pi} \frac{2x \sin x}{1+\cos^2 x}\, dx = 2 \cdot \frac{\pi^2}{2} = \pi^2$$ Therefore, the value of the definite integral is $\boxed{\pi^2}$, which corresponds to **Option 3**.