How can $\int \frac{2^x3^xdx}{9^x-4^x}$ be found?

Question

I am stuck on this question: $\int \frac{2^x3^xdx}{9^x-4^x}$. I have tried to solve it by writing it down as $\int \frac{2^x3^xdx}{(3^x-2^x)(3^x+2^x)}$ and making some substitution, but I still can't find the solution. Could you please suggest any hints or methods?

Answer 1

HINT:

$$\dfrac{2^x3^x}{9^x-4^x}=\dfrac1{(3/2)^x-(2/3)^x}$$

Choose $(3/2)^x$ OR $(2/3)^x=u$

Utilize the fact $(3/2)^x\cdot(2/3)^x=1$

Answer 2

HINT: divide both numerator and denominator by $\displaystyle 2^x3^x$ and then substitute $$u= \frac{3^x}{2^x}$$