Answer
The formulas are equivalent.
Work Step by Step
First, we manipulate $(\sec x+\tan x)$ by multiplying it with 1...
... and for 1, we choose $\displaystyle \frac{\sec x-\tan x}{\sec x-\tan x}$.
$\displaystyle \sec x+\tan x=(\sec x+\tan x)\cdot\frac{\sec x-\tan x}{\sec x-\tan x} =\frac{\sec^{2}x-\tan^{2}x}{\sec x-\tan x}$
... Pythagorean identity: $1+\tan^{2}=\sec^{2}$ ...
so, $\sec^{2}x-\tan^{2}x=1$
and
$\displaystyle \sec x+\tan x=\frac{1}{\sec x-\tan x} =(\sec x-\tan x)^{-1}$
Now, since the two LHS's are equal, we check if the RHS's are also equal.
The RHS of formula 1 =
$\ln(\sec x+\tan x)+C =\ln(\sec x-\tan x)^{-1}+C$
...Log Property: $\ln M^{n}=n\ln M...$
$=(-1)\ln(\sec x-\tan x)+C$
$... $= The RHS of formula 2
Therefore, the formulas are equivalent.