Simplifying a Trigonometric Expression I have to prove that:
$$x \sec x - \ln |\sec x + \tan x| + C$$
is the indefinite integral of:
$$x \sec x \tan x $$ 
by taking the derivative.
I've got far enough to get:
$$x\sec x\tan x + \sec x -\dfrac{|\sec x+\tan x|(\sec^2 x + \sec x \tan x)}{|\sec x + \tan x|}.$$   
Kind of stuck here.  Am I able to cancel out the $|\sec x + \tan x|$ on top and bottom and then set $-\sec x$ equal to $\sec^2 x + \sec x \tan x$?  I'm guessing that's not right though.
Sorry for the crummy way I have it setup, feel free to edit it.

Two issues—first, as suggested in Jerry's answer, you have a factor of $|\sec x+\tan x|$ in the numerator of the last term of your derivative that does not belong there.  Second, the derivative of $\ln|x|$ is $\frac{1}{x}$ (no absolute value), so $\frac{d}{dx}\ln|f(x)|=\frac{1}{f(x)}\cdot f'(x)$ with the chain rule.
