$\int {e}^{2x}=1\frac{l}{2}\int 2{e}^{2x}dx$
If $2dx=d\left(2x\right)$, 2 is the derivative of 2x
And we have:
$\frac{1}{2}\int {e}^{2x}d\left(2x\right)$
Let u=2x
Thus, $\frac{1}{2}\int {e}^{u}du=\frac{1}{2}{e}^{u}=\frac{1}{2}{e}^{2x}$

Annie Levasseur

Beginner2021-12-15Added 30 answers

You can use the technique of integration by substitution to find this. $f\left(x\right)=\frac{1}{2}{e}^{2x}$ has $f}^{\prime}\left(x\right)={e}^{2x$, so it's an antiderivative. And the general one is: $\frac{1}{2}{e}^{2x}+C$