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.

Your derivative for $x\sec\;x$ is correct; for the second term, note that $\frac{\mathrm d}{\mathrm dx}\ln(f(x))=\frac{f^{\prime}(x)}{f(x)}$ . Apply the formula accordingly.
