# How do you find the antiderivative of e^x(1-(e^-x)sec^2x) dx?

Nov 15, 2017

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

#### Explanation:

Simplify the expression:

${e}^{x} \left(1 - {e}^{- x} {\sec}^{2} x\right) = {e}^{x} - {\sec}^{2} x$

then using the linearity of the integral:

$\int {e}^{x} \left(1 - {e}^{- x} {\sec}^{2} x\right) \mathrm{dx} = \int {e}^{x} \mathrm{dx} - \int {\sec}^{2} x \mathrm{dx}$

Both terms are standard integrals:

$\int {e}^{x} \mathrm{dx} = {e}^{x} + {c}_{1}$

$\int {\sec}^{2} x \mathrm{dx} = \tan x + {c}_{2}$

and then:

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