Answer
$\displaystyle\int{\left( 1+\tan x \right) ^2\sec x\ dx}=\frac{1}{2}\sec x\tan x+\frac{1}{2}\ln \left| \tan x+\sec x \right|+2\sec x+C
$
Work Step by Step
$
I=\displaystyle\int{\left( 1+\tan x \right) ^2\sec x\ dx}\\
I=\displaystyle\int{\left( 1+\tan ^2x+2\tan x \right) \sec x\ dx}\\
I=\displaystyle\int{\left( \sec ^2x+2\tan x \right) \sec x\ dx}\\
I=\displaystyle\int{\sec ^3x+2\tan x\sec x\ dx}\\
I=\displaystyle\int{\sec ^3x\ dx}+2\int{\tan x\sec x\ dx}$
$2\displaystyle\int{\tan x\sec x\,\,dx}\\
2\sec x$
$\displaystyle\int{\sec ^3x\ dx}\\
\displaystyle\int{\sec ^2x\sec x\ dx}
$
$
\begin{matrix}
u=\sec x& du=\sec x\tan x\\
dv=\sec ^2x& v=\tan x\\
\end{matrix}
$
$
\displaystyle\int{\sec ^3x\,\,dx}=\sec x\tan x-\int{\sec x\tan ^2x\ dx}\\
\displaystyle\int{\sec ^3x\,\,dx}=\sec x\tan x-\int{\sec x\left( \sec ^2x-1 \right) \,\,dx}\\
\displaystyle\int{\sec ^3x\,\,dx}=\sec x\tan x-\int{\sec ^3x-\sec x\,\,dx}\\
\displaystyle\int{\sec ^3x\,\,dx}=\sec x\tan x-\int{\sec ^3x\ dx+\int{\sec x\ dx}}\\
2\displaystyle\int{\sec ^3x\,\,dx}=\sec x\tan x+\int{\sec x\ dx}\\
2\displaystyle\int{\sec ^3x\,\,dx}=\sec x\tan x+\ln \left| \tan x+\sec x \right|\\
\displaystyle\int{\sec ^3x\,\,dx}=\frac{1}{2}\sec x\tan x+\frac{1}{2}\ln \left| \tan x+\sec x \right|$
$I=\frac{1}{2}\sec x\tan x+\frac{1}{2}\ln \left| \tan x+\sec x \right|+2\sec x+C
$
