Is there a direct proof for $\int_0^{2\pi}\frac{r^2+r(\cos t-\sin t)}{1+2r\cos t+r^2}dt=2\pi$

Utilize the Fourier series

$$\frac{r^2-1}{1+2r\cos t+r^2}=1+ 2\sum_{n=1}^{\infty}\frac{(-1)^n}{r^n}\cos(nt ) $$ and observe that only the first two terms in the series survive the integration, i.e.

$$\int_0^{2\pi}\frac{r^2+r(\cos t-\sin t)}{1+2r\cos t+r^2}dt = \frac1{r^2-1}\int_0^{2\pi}(r^2- 2\cos^2t)dt =2\pi$$

We can reduce the problem a little bit. By elementary algebra and symmetry you'll see that it suffices to show that $$ \int_0^{2 \pi} \dfrac{1+r \cos t}{1+2 r \cos t + r^2} \,dt = 0. $$

Under suitable transformation we can evaluate this integral as $$ \left[ -\frac{1}{2} \arctan \left(\frac{(r+1) \cos \left(\frac{t}{2}\right)}{\sin \left(\frac{t}{2}\right)-r \sin \left(\frac{t}{2}\right)}\right)+\frac{1}{2} \arctan \left(\frac{(r+1) \cos \left(\frac{t}{2}\right)}{r \sin \left(\frac{t}{2}\right)-\sin \left(\frac{t}{2}\right)}\right)+\frac{t}{2}\right]_0^{2 \pi} = 0 $$

note: Out of laziness I computed the primitive with Wolfram Mathematica, but the transformation is standard.