im not sure whether im right but hopefully it helps.
\int \frac{(sec^2 x)}{(3 + 4tan^2 x)} dx \\\\ =\int \frac{1}{(3 + 4tan^2 x)} d(tan x)\\\\=\frac{1}{4}\int \frac{1}{(\frac{3}{4} + tan^2 x)} d(tan x)\\\\=\frac{2}{\sqrt{3}}.\frac{1}{4}.\tan^{-1}(\frac{2tanx}{\sqrt{3}})+C