Question

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

Answer 1

`e^(x+3)+e^(x-3)+C`

Explanation:

Given: `inte^(x+3)+e^(x-3) \ dx`

`=inte^(x+3) \ dx+inte^(x-3) \ dx`

Let `u=x+3,:.du=dx`

`=inte^u \ du+inte^(x-3) \ dx`

Let `z=x-3,:.dz=dx`

`=inte^u \ du+inte^z \ 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`