A simple integration

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,

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

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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This problem is a duplicate of this one and you can see there that this problem can also be solved using IBP but this approach is much more elementary and easy.

I posted my solution (this one) as a comment there as you can see if you open that problem and since nobody has posted a solution here, I'm copying my comment from there and posting it as a solution here. :)

$I=\int_{-π}^{π} \dfrac{2x+2x \sin x}{1+\cos^2x} dx \\ = \int_{-π} ^{π} \dfrac{2x}{1+\cos^2x}dx + \int_{-π}^{π} \dfrac{2x \sin x} {1+\cos^2x} dx$ But the function $\dfrac{2x}{1+\cos^2x}$ is odd thus its integral is equal to zero and the function $\dfrac{2x \sin x}{1+\cos^2x}$ is even.

Therefore $I=0+\int_{-π} ^{π} \dfrac{2x \sin x}{1+\cos^2x} dx \\ = 2\int_{0}^{π}\dfrac{2x \sin x}{1+\cos^2x} dx \\ \Rightarrow I = 4\int_{0}^{π} \dfrac{x \sin x}{1+\cos^2x} dx$

Define $\cos^{-1}:[-1,1]\rightarrow [0,π]$ which is the inverse function of $\cos:[0, π] \rightarrow [-1,1]$

Let $x=\cos^{-1} u \Leftrightarrow u=\cos x \Leftrightarrow du=-\sin xdx$ Then it is $\cos0=1$ and $\cosπ=-1$ .Hence

$I=4\int_{-1}^{1} \dfrac{\cos^{-1}u}{u^2+1}du$

We know that the function $\sin:[-\frac{π}{2},\frac{π}{2}]$ is odd thus

$\sin x = - \sin(-x), \forall x \in [-\frac{π}{2},\frac{π}{2}] \\ \Rightarrow \cos(\frac{π}{2}+x)=-\cos(\frac{π}{2}-x), \forall x \in [-\frac{π}{2},\frac{π}{2}]$

Thus the function $f(x) = \cos(\frac{π}{2}-x), \forall x \in [-\frac{π}{2},\frac{π}{2}]$ is odd. Let $g:[-1,1]\rightarrow [-\frac{π}{2},\frac{π}{2}]$ be the inverse function of $f$ then $g$ is also odd. And it holds that

$g(\cos(\frac{π}{2}-x)) = x, \forall x \in [-\frac{π}{2},\frac{π}{2}] \\ \Rightarrow g(\cos(x)) =-x+\dfrac{π}{2}, \forall x \in [0,π] \\ \Rightarrow g(x)=-\cos^{-1}x+\dfrac{π}{2}, \forall x \in [-1,1]$

Therefore the integral equals to:

$I=-4\int_{-1}^{1} \dfrac{g(u)}{u^2+1}du + 4\int_{-1}^{1} \dfrac{\frac{π}{2}}{u^2+1}du$

But the function $\dfrac{g(x)}{x^2+1}$ is odd thus its integral is equal to zero. Lastly the given integral equals to:

$I=2π \int_{-1}^{1}\dfrac{1}{u^2+1}du \\ = 2π[ \arctan u]_{-1}^{1} \\ = 2π(\frac{π}{4}-(-\frac{π}{4})) = \boxed{π^2}$