# How do you find the indefinite integral of int (e^(x+3)+e^(x-3))dx?

Apr 14, 2018

${e}^{x + 3} + {e}^{x - 3} + C$

#### Explanation:

Given: $\int {e}^{x + 3} + {e}^{x - 3} \setminus \mathrm{dx}$

$= \int {e}^{x + 3} \setminus \mathrm{dx} + \int {e}^{x - 3} \setminus \mathrm{dx}$

Let $u = x + 3 , \therefore \mathrm{du} = \mathrm{dx}$

$= \int {e}^{u} \setminus \mathrm{du} + \int {e}^{x - 3} \setminus \mathrm{dx}$

Let $z = x - 3 , \therefore \mathrm{dz} = \mathrm{dx}$

$= \int {e}^{u} \setminus \mathrm{du} + \int {e}^{z} \setminus \mathrm{dz}$

$= {e}^{u} + {e}^{z}$

Substitute back $u = x + 3 , z = x - 3$ to get:

$= {e}^{x + 3} + {e}^{x - 3}$

Add a constant at the end.

$= {e}^{x + 3} + {e}^{x - 3} + C$