Answer
$$x\sec x -\ln| \sec x +\tan x|+c$$
Work Step by Step
Given $$\int x\sec x\tan xdx$$
Let
\begin{align*}
u&=x\ \ \ \ \ \ \ \ \ \ \ \ \ \ dv=\sec x\tan xdx\\
du&=dx\ \ \ \ \ \ \ \ \ \ \ \ \ \ v=\sec x
\end{align*}
Then
\begin{align*}
\int x\sec x\tan xdx&=x\sec x -\int \sec x dx\\
&=x\sec x -\ln| \sec x +\tan x|+c
\end{align*}
