Answer
$3 \ln |\sec (2x-4)+\tan (2x-4) | +C$
Work Step by Step
Our aim is to solve the integral $ \int 6 \sec (2x-4) \ dx$
Let us consider that $a =(2x-4)$ and $\dfrac{da}{dx}=2 \implies dx=\dfrac{1}{2} \ da$
Now, $ \int 6 \sec (2x-4) \ dx = \int 6 \sec a \times (\dfrac{1}{2}) \ da$
or, $= 3 \int \sec a \ da$
or, $=3 \ln |\sec a+\tan a | +C$
or, $=3 \ln |\sec (2x-4)+\tan (2x-4) | +C$
