Simplify:
$\int \frac{secx}{secx+tanx}$

$\int \frac{\sec xdx}{\sec x+\tan x}$

$=\int \frac{\sec xdx}{\sec x+\tan x}\times \frac{\sec x-\tan x}{\sec x-\tan x}$

$=\int \frac{\sec x(\sec x-\tan x)dx}{{\sec }^{2}x-{\tan }^{2}x}$

$=\int \sec x(\sec x-\tan x)dx$ since ${\sec }^{2}x-{\tan }^{2}x=1$

$=\int {\sec }^{2}xdx-\int \sec x\tan xdx$

$=\tan x-\sec x+c$ since $\int {\sec }^{2}xdx=\tan x+c$ and $\int \sec x\tan xdx=\sec x+c$