Answer
$$ = \frac{1}{2}\sinh \left( {2x - 3} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\cosh \left( {2x - 3} \right)} dx \cr
& {\text{substitute }}u = 2x - 3,{\text{ }}du = 2dx \cr
& = \int {\cosh \left( {2x - 3} \right)} dx = \frac{1}{2}\int {\cosh udu} \cr
& {\text{find the antiderivarive using the theorem 6}}{\text{.8}}{\text{.3 }}\left( {{\text{see page 476}}} \right) \cr
& = \frac{1}{2}\sinh u + C \cr
& {\text{write in terms of }}x \cr
& = \frac{1}{2}\sinh \left( {2x - 3} \right) + C \cr} $$
