Answer
$2 \sqrt {\ln (\sec x +\tan x)}+c$
Work Step by Step
Solve $\int\dfrac{\sec xdx}{\sqrt {\ln (\sec x +\tan x)}}$
Let $p=\ln (\sec x +\tan x)$ and $dp=\sec xdx$
This implies $\int\dfrac{\sec xdx}{\sqrt {\ln (\sec x +\tan x)}}=\int\dfrac{dp}{\sqrt p}=2p^{1/2}+c=2 \sqrt p+c$
Since, $p=\ln (\sec x +\tan x)$
Thus, $\int\dfrac{\sec xdx}{\sqrt {\ln (\sec x +\tan x)}}=2 \sqrt {\ln (\sec x +\tan x)}+c$
