Nguyên hàm 2

Tính nguyên hàm: $I=\displaystyle\int \frac{\sin x-\sqrt{3}\cos x+2}{\sqrt{3}\sin x+\cos x+2}dx$

$$I=-\int \dfrac{\sqrt 3 \cos x -\sin x }{\sqrt 3 \sin x +\cos x +2}dx -2\int \dfrac{dx}{\sqrt 3 \sin x +\cos x +2}$$ $$I=-\int \dfrac{d\left(\sqrt 3 \sin x +\cos x +2\right)}{\sqrt 3 \sin x +\cos x +2}-\int \dfrac{dx}{\dfrac{\sqrt 3}{2} \sin x +\dfrac{1}{2} \cos x +1}$$ $$I=-\ln \left|\sqrt 3 \sin x +\cos x +2\right| -\int \dfrac{dx}{\cos \left(x-\dfrac{\pi}{3}\right) +1}$$ $$I=-\ln \left|\sqrt 3 \sin x +\cos x +2\right| -\int \dfrac{dx}{2\cos^2 \left(\dfrac{x}{2}-\dfrac{\pi}{6}\right)}$$ $$I =-\ln \left|\sqrt 3 \sin x +\cos x +2\right| -\int \dfrac{d\left(\dfrac{x}{2}-\dfrac{\pi}{6}\right)}{\cos^2 \left(\dfrac{x}{2}-\dfrac{\pi}{6}\right)} $$ $$I =-\ln \left|\sqrt 3 \sin x +\cos x +2\right| -\tan\left(\dfrac{x}{2}-\dfrac{\pi}{6}\right) +C$$