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\\
u&= dx\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ dv= \sec x
\end{align*}
then
\begin{align*}
\int x \sec x \tan x d x&=x \sec x-\int \sec x d x\\
&=x \sec x-\ln |\sec x+\tan x|+C
\end{align*}
