# How do you integrate  (e^(2x-1))-1?

$= \frac{1}{2} {e}^{2 x - 1} - x + C$
$\int \setminus \left({e}^{2 x - 1}\right) - 1 \setminus \mathrm{dx}$
$= \int \setminus \frac{d}{\mathrm{dx}} \left(\frac{1}{2} {e}^{2 x - 1}\right) - 1 \setminus \mathrm{dx}$
$= \frac{1}{2} {e}^{2 x - 1} - x + C$