# How to integrate? int 2e^x+2e^-x dx

Nov 28, 2017

$2 \left({e}^{x} - {e}^{- x}\right) + c$

#### Explanation:

I don't see the "dx" in there, I'm going to assume your problem as given to you has it.

$\int \left(2 {e}^{x} + 2 {e}^{- x}\right) \mathrm{dx} = 2 \int {e}^{x} \mathrm{dx} + 2 \int {e}^{- x} \mathrm{dx}$

...for the first term, you know that the antiderivative of ${e}^{x} = {e}^{x}$

for the second term, remember that $\int \left({e}^{u} \mathrm{du}\right) = {e}^{u}$. In this case, $u = - x$, so du = -1dx. So you can convert the second term to a form you can readily integrate by multiplying by $\left(- 1\right) \left(- 1\right)$, which turns your second integral term into:

$- 2 \int {e}^{- x} \left(- \mathrm{dx}\right)$

so, your original problem's integral evaluates to:

$2 {e}^{x} - 2 {e}^{- x} + c$

$= 2 \left({e}^{x} - {e}^{- x}\right) + c$

GOOD LUCK