Not Trigonometric Substitution

Let me present an alternative solution that doesn't use IBP and instead uses simple trigonometric substitution later on. We use a particular property of definite integrals, namely: $\displaystyle\int\limits_a^b f(x)\,dx=\int\limits_a^b f(a+b-x)\,dx$

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Moving on to the problem, consider the integral as $I$ . Using the property mentioned above, we have,

$I=\int\limits_{-\pi}^\pi \frac{2x(1+\sin x)}{1+\cos^2 x}\,dx=-\int\limits_{-\pi}^\pi \frac{2x(1-\sin x)}{1+\cos^2 x}\,dx\\ \implies 2I=\int\limits_{-\pi}^\pi \frac{2x(1+\sin x -1+\sin x)}{1+\cos^2 x}\,dx\\ \implies 2I=\int\limits_{-\pi}^\pi \frac{4x\sin x}{1+\cos^2 x}\,dx$

Now, this integrand is even and hence, we can simplify it as,

$2I=2\cdot \int\limits_0^\pi \frac{4x\sin x}{1+\cos^2 x}\,dx\implies I=\int\limits_0^\pi \frac{4x\sin x}{1+\cos^2 x}\,dx$

Using the property mentioned at the very beginning, the integral can be written as,

$I=\int\limits_0^\pi \frac{4\pi\sin x}{1+\cos^2 x}-I\implies 2I=4\pi\cdot \int\limits_0^\pi \frac{\sin x}{1+\cos^2}\,dx$

Make the substitution $t=(-\cos x)$ and $\,dt=\sin x \,dx$ to get,

$2I=4\pi\cdot\int\limits_{-1}^1 \frac{\,dt}{1+t^2}$

This integrand is again even and also we can easily identify the integrand to be the derivative of $\arctan t$ . Hence, we have,

$2I=8\pi\cdot\left[\arctan t\right]_0^1=8\pi\left(\frac{\pi}{4}-0\right)\\ \implies \boxed{I=~\pi^2}$

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@kritarth lohomi , Problem? :3

The limits are from $\pi$ to $\pi$ . Isn't the answer zero?