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

Aug 3, 2016

$= - {e}^{-} \left(x + 1\right) + C$

#### Explanation:

for starters

${e}^{x} / {e}^{2 x + 1} = {e}^{x} \cdot {e}^{-} \left(2 x + 1\right) = {e}^{-} \left(x + 1\right)$

and as $\frac{d}{\mathrm{dx}} {e}^{f \left(x\right)} = f ' \left(x\right) {e}^{f \left(x\right)}$

you are therefore looking at

$\int \setminus {e}^{-} \left(x + 1\right) \setminus \mathrm{dx}$

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

$= - {e}^{-} \left(x + 1\right) + C$

of course, for $\int \setminus {e}^{-} \left(x + 1\right) \setminus \mathrm{dx}$

you can introduce a sub like $u = - \left(x + 1\right) , \mathrm{du} = - \mathrm{dx}$ and solve it as

$\int \setminus - {e}^{u} \setminus \mathrm{du} = - {e}^{u} + C = - {e}^{-} \left(x + 1\right) + C$

if you prefer.