Integral $\int\frac{6^x}{9^x-4^x}dx $

Divide top and bottom by $6^x$.

So we want to integrate $\frac{1}{a^x-a^{-x}}$, where $a=3/2$.

This is a close relative of the integral of $\frac{1}{\sinh t}$. To integrate that, we multiply top and bottom by $\sinh t$, and use the fact that $\sinh^2 t=\cosh^2 t-1$.

Now if we feel like it we can rewrite without mentioning $\sinh$ and $\cosh$.

Added: @Fly by Night mentions in a comment that according to various pieces of software, the function has no known antiderivative, and asks for detail. So we do the details of the $\sinh$ and $\cosh$ approach, as promised not using these functions explicitly.

So we want $\int \frac{dx}{a^x-a^{-x}}$. Multiply top and bottom by $a^x-a^{-x}$, and note that $(a^x-a^{-x})^2=(a^x+a^{-x})^2 -4$. (This is essentially the $\sinh^2 t=\cosh^2 t-1$ identity.)

So we want to find $$\int \frac{(a^x-a^{-x})\,dx}{(a^x+a^{-x})^2-4}.$$ Let $u=a^x+a^{-x}$. Then our integral becomes $$\int \frac{du}{u^2-4},$$ which I expect can be handled by Wolfram Alpha.

Divide top and bottom by $ 4 ^ x $. We have: $$\int\dfrac{\left(\frac{3}{2}\right)^x}{\left(\frac{9}{4}\right)^x-1}dx=\int\dfrac{\left(\frac{3}{2}\right)^x}{\left(\left(\frac{3}{2}\right)^x\right)^2-1}dx$$ is replaced: $$u=\left(\frac{3}{2}\right)^x$$ $$du=\left(\frac{3}{2}\right)^x\cdot\ln\left(\frac{3}{2}\right)dx$$ replacing in the integral: $$I=\dfrac{1}{\ln\left(\frac{3}{2}\right)}\int\dfrac{1}{u^2-1}du$$ solving: $$I=\dfrac{1}{\ln\left(\frac{3}{2}\right)}\cdot\dfrac{1}{2}\int\dfrac{2}{u^2-1}du=\dfrac{1}{2\ln\left(\frac{3}{2}\right)}\int\left(\dfrac{1}{u-1}-\dfrac{1}{u+1}\right)du$$ $$I=\dfrac{1}{2\ln\left(\frac{3}{2}\right)}\left(\ln(u-1)-\ln(u+1)\right)=\dfrac{1}{2\ln\left(\frac{3}{2}\right)}\ln\left(\dfrac{u-1}{u+1}\right)$$

Finally: $$\int\dfrac{6^x}{9^x-4^x}dx=\dfrac{1}{2\ln\left(\frac{3}{2}\right)}\ln\left\vert\dfrac{\left(\frac{3}{2}\right)^x-1}{\left(\frac{3}{2}\right)^x+1}\right\vert+C$$ if you want you can multiply top and bottom within the logarithm, by $ 2 ^ x $. $$\int\dfrac{6^x}{9^x-4^x}dx=\dfrac{1}{2\ln\left(\frac{3}{2}\right)}\ln\left\vert{\dfrac{3^x-2^x}{3^x+2^x}}\right\vert+C$$

Note that $$\frac{1}4\cdot\frac{3^x+2^x}{3^x-2^x}-\frac{1}4\cdot\frac{3^x-2^x}{3^x+2^x}$$ is the integrand. Now use what @J.H. suggested.