$ \int \sec x . \sec^2 x dx$
=> $ \int \sqrt{1+\tan ^2 x}. \sec^2. x dx$
$\tan x =t=> \sec^2 x dx=dt$
=> $\int \sqrt {1+t^2}.dt$
=> $\large\frac{1}{2}$$ t \sqrt {1+t^2} + \large\frac{1}{2}$$ \log \{ t+ \sqrt {1+t^2}\}+c$
=> $\large\frac{1}{2}$$ \tan x . \sec x +\large\frac{1}{2} $$ \log (\tan x + \sec x \}+c$
Hence b is the correct answer.