Question:

Divide numerator and denominator by $\cos^2 x$<br>$$I = \int \frac{\sec^2 x dx}{2 \sec^2 x + \tan^2 x} = \int \frac{\sec^2 x dx}{2 + 3 \tan^2 x}$$<br>Let $\sqrt{3} \tan x = t \Rightarrow \sqrt{3} \sec^2 x dx = dt$<br>So $$I = \frac{1}{\sqrt{3}} \int \frac{dt}{2 + t^2} = \frac{1}{\sqrt{3}} \cdot \frac{1}{\sqrt{2}} \tan^{-1} \frac{t}{\sqrt{2}} + C = \frac{1}{\sqrt{6}} \tan^{-1} \left( \frac{\sqrt{3} \tan x}{\sqrt{2}} \right) + C$$