What is the integral of e^(0.5x)?

$2 {e}^{0.5 x} + C$

Explanation:

$\setminus \int {e}^{0.5 x} \setminus \mathrm{dx}$

$= \setminus \int {e}^{0.5 x} \frac{1}{0.5} d \left(0.5 x\right)$

$= \frac{1}{0.5} \setminus \int {e}^{0.5 x} \setminus d \left(0.5 x\right)$

$= 2 {e}^{0.5 x} + C$

Jul 18, 2018

$2 {e}^{\frac{1}{2} x} + C$
Substiting $t = \frac{1}{2} x$ we get
$x = 2 t$ so $\mathrm{dx} = 2 \mathrm{dt}$ and our integral is given by $2 \int {e}^{t} \mathrm{dt} = 2 {e}^{t} + C = 2 {e}^{\frac{1}{2} x} + C$