Question

integral sec^2x/3+tan x dx​

Answer 1

$\mathsf{u = 3 + tg(x) \to \: du = {sec}^{2}(x)dx} \\ \displaystyle\mathsf{\int\dfrac{{sec}^{2}(x)}{3+tg(x)}dx} \\ = \displaystyle\mathsf{\int\dfrac{du}{u} \: du = ln|u| + k }$

$\displaystyle\mathsf{\int\dfrac{{sec}^{2}(x)}{3+tg(x)}dx=ln|3+tg(x)|+k}$