I'm currently teaching myself calculus and I'm probably trying to run before I can walk, but I've been working on this problem..
I managed to find the correct result for:
$$\int_{0}^{\infty }(2e^{-3x}+4e^{-7x})^2dx$$
by expanding it to:
$$\int_{0}^{\infty}4e^{-6x}+16e^{-10x}+16e^{-14x}dx$$
and then working from there.
Is there a better/more general approach I could have taken? I've attempted to solve it using substitution but haven't had any success...