Find The Area Of The Region:

$\displaystyle y = \frac{\pi}{4x}$ and $\displaystyle y = tan x$ between $\displaystyle x = \frac{\pi}{6}$ and the first intersection point of the curves to the right of $\displaystyle x = 0$

At the moment I have the graph drawn and I need to find the first intersection point to the right of $\displaystyle x = 0$

I also have the integral wrote up along with the antiderivative so I hope its right.

$\displaystyle A = \int_{\frac{\pi}{6}} \left( (tan x) - (\frac{\pi}{4x}) \right) dx$

$\displaystyle A = -ln ( cos(x) ) - \frac{\pi}{4} ln x \mid_{\frac{\pi}{6}}$

All I need is the right value and im set

thanks to anyone who helps : )