## Calculus, 10th Edition (Anton)

$=\frac{1}{2}e^{2x}+C$
Using a u-substitution of $u=2x$ leaves a $du=2dx$, meaning that $\frac{1}{2}du=dx$. This yields $$\frac{1}{2}\int{e^{u}}du=\frac{1}{2}e^{u}+C$$ Changing back to terms of $u$ yields $\frac{1}{2}e^{2x}+C.$