Answer
$\displaystyle \frac{1}{2}\sinh 2x+C$
Work Step by Step
Use Th.5.18, Derivatives and Integrals of Hyperbolic Functions
$\displaystyle \int\cosh udu=\sinh u+C$
----
$\displaystyle \int\cosh 2xdx=\left[\begin{array}{lll}
t=2x & & \\
dt=2dx & \Rightarrow & \frac{1}{2}dt=dx
\end{array}\right]$
$=\displaystyle \frac{1}{2}\int\cosh(t)dt$
$=\displaystyle \frac{1}{2}\sinh t+C$
$=\displaystyle \frac{1}{2}\sinh 2x+C$
