The integral $$\displaystyle\int \dfrac{1 - \dfrac{1}{\sqrt{3}}(\cos x - \sin x)}{1 + \dfrac{2}{\sqrt{3}}\sin 2x} dx$$ is equal to

Given,

$$I=\int \frac{1-\frac1{\sqrt3}(\cos x-\sin x)}{1+\frac2{\sqrt3}\sin2x}\,dx$$

Multiply numerator and denominator by

$$\frac{\sqrt3}{2}:$$

$$I=\int \frac{\frac{\sqrt3}{2}-\frac12(\cos x-\sin x)}{\frac{\sqrt3}{2}+\sin2x}\,dx$$

Using

$$\sin2x=2\sin x\cos x,$$

$$I=\int \frac{\frac{\sqrt3}{2}-\frac12\cos x+\frac12\sin x}{2\sin x\cos x+\frac{\sqrt3}{2}}\,dx$$

Now write denominator as

$$2\sin x\cos x+\frac{\sqrt3}{2}=2\left(\sin x+\frac12\right)\left(\cos x+\frac{\sqrt3}{2}\right)$$

Hence,

$$I=\int \frac{\frac{\sqrt3}{2}\cos x-\frac12\cos x-\frac{\sqrt3}{2}\sin x+\frac12\sin x}{2\left(\sin x+\frac12\right)\left(\cos x+\frac{\sqrt3}{2}\right)}\,dx$$

Split the fraction:

$$I=\frac12\int\frac{\cos x-\frac1{\sqrt3}\sin x}{\sin x+\frac12}\,dx-\frac12\int\frac{\sin x-\frac1{\sqrt3}\cos x}{\cos x+\frac{\sqrt3}{2}}\,dx$$

Now use substitutions.

For the first integral, let

$$u=\sin x+\frac12$$

Then,

$$du=\cos x\,dx$$

Hence,

$$\frac12\int\frac{du}{u}=\frac12\ln|u|$$

For the second integral, let

$$v=\cos x+\frac{\sqrt3}{2}$$

Then,

$$dv=-\sin x\,dx$$

Hence,

$$-\frac12\int\frac{dv}{v}=-\frac12\ln|v|$$

Therefore,

$$I=\frac12\ln\left|\sin x+\frac12\right|-\frac12\ln\left|\cos x+\frac{\sqrt3}{2}\right|+C$$

Combining logarithms,

$$I=\frac12\ln\left|\frac{\sin x+\frac12}{\cos x+\frac{\sqrt3}{2}}\right|+C$$

Using

$$\sin x=\frac{2\tan\frac x2}{1+\tan^2\frac x2},\qquad\cos x=\frac{1-\tan^2\frac x2}{1+\tan^2\frac x2},$$

this simplifies to

$$I=\frac12\ln\left|\frac{\tan\frac x2+\frac1{2}}{\tan\frac x2+\frac{\sqrt3}{2}}\right|+C$$

Hence,

$$\boxed{I=\frac12\ln\left|\frac{\tan\frac x2+\frac1{2}}{\tan\frac x2+\frac{\sqrt3}{2}}\right|+C}$$