How to Integrate 2/e3x(6+e-3x)dx Using Substitution

[Iskander]

Can't integrate the "e"s here

Homework Statement

[tex]\int 2/e^3x(6+e^-3x)dx[/tex]

Homework Equations

I have no idea as to what equations to use, other than $$\int$$ e[tex]^u[/tex] du = e [tex]^u[/tex] + C

The Attempt at a Solution

$$\int$$ 2/(6e<sup>3x +1) dx
u = 6e3x +1
du = 18e3x dx

and that's as far as I can get, I can't think of what else to do. HELP!

Thanks in advance.</sup>

 

[Mu naught]

please fix your latex code to make it readable

 

[eumyang]

Is it supposed to be this?
$$\int \frac{2}{e^{3x}} (6 + e^{-3x}) dx$$

Or this?
$$\int \frac{2}{e^{3x} (6 + e^{-3x})} dx$$

If it's the second, then I wouldn't distribute the e^(3x) at all. I would move it to the numerator, and then use substitution: u = (6 + e^(-3x)). Try it and see what happens.