## Calculus (3rd Edition)

$$12\sec{x}+C$$
$\int 12\sec{x}\tan{x} dx$ Pulling 12 out of the integral: $12\int\sec{x}\tan{x} dx$ Given $$\int\sec{x}\tan{x} dx=\sec{x}+C$$ $12\int\sec{x}\tan{x} dx$ $=12\sec{x}+C$