<s>[INST] Solve $\int{\frac{x-1}{x\sqrt{3x^2-4}}}\;dx$ In solving the definite integral $$\int{\frac{x-1}{x\sqrt{3x^2-4}}}\;dx$$I tried to do $$\int{\frac{x-1}{x\sqrt{3x^2-4}}}\;dx=\int{\frac{1}{\sqrt{3x^2-4}}}\;dx-\int{\frac{1}{x\sqrt{3x^2-4}}}\;dx\Longrightarrow$$ If $x=\frac{2}{\sqrt{3}}\sec(u)\Longrightarrow dx=\frac2{\sqrt3}\tan(u)\sec(u)\;du$, then
$$\int{\frac{1}{\sqrt{3x^2-4}}}\;dx-\int{\frac{1}{x\sqrt{3x^2-4}}}\;dx=$$ $$\frac{2}{\sqrt3}\int\frac{\tan(u)\sec(u)}{2\sqrt{\left(\frac{\sqrt3}{2}\sec(u)\right)^2-1}}\;du-\frac1{2}\int\frac{\frac{\sqrt3}{2}}{\frac{\sqrt3}{2}x\sqrt{\left(\frac{\sqrt3}{2}x\right)^2-1}}\;dx=$$ $$\frac{1}{\sqrt3}\int\sec{u}\;du-\frac12\sec^{-1}\left(\frac{\sqrt3}{2}x\right)$$
In my feedback says that the result is $$\frac{1}{\sqrt3}\log\left|\sqrt3x+\sqrt{3x^2-4}\right|-\frac12\sec^{-1}{\left(\frac{\sqrt3}{2}x\right)}+C$$
It also says that I have to use this method of trigonometric substitution.
The path is this? 
I know $\int\sec(t)dt=\log|\sec t+\tan t|$
 [/INST] You are essentially finished. We have $\sec u=\frac{\sqrt{3}x}{2}$ and $\tan u=\frac{\sqrt{3x^2-4}}{2}$. Thus your procedure gives as first term 
$$\frac{1}{\sqrt{3}}\log\left|\frac{\sqrt{3}x+\sqrt{3x^2-4}}{2}\right|.$$
This is equal to 
$$\frac{1}{\sqrt{3}}\log\left|\sqrt{3}x+\sqrt{3x^2-4}\right|-\frac{1}{\sqrt{3}}\log 2.$$
So the two versions differ by a constant, which can be absorbed in the constant of integration. 
</s>