TITLE: Integral $\int\frac{6^x}{9^x-4^x}dx $
QUESTION [8 upvotes]: How to solve this integral:
$$\int\frac{6^x}{9^x-4^x}dx $$
(I notice that $\frac{6^x}{9^x-4^x}=\frac{2^x3^x}{(3^x-2^x)(3^x+2^x)}$)
Thank you!

REPLY [3 votes]: 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 $.
$$\int\dfrac{6^x}{9^x-4^x}dx=\dfrac{1}{2\ln\left(\frac{3}{2}\right)}\ln\left\vert{\dfrac{3^x-2^x}{3^x+2^x}}\right\vert+C$$