We need to get this $\frac{\ln{x}}{x^3-9x}= \sum_{k=0}^{\infty} a_k (2x-3)^k$
So I am doing the next steps:
As we know everything start with this, call it eq(1): $$\frac{1}{1-x}= \sum_{k=0}^{\infty} (x)^k$$
Doing some algebra: $$\frac{1}{t}=\frac{1}{t+\frac{3}{2}-\frac{3}{2}}=\frac{1}{\frac{3}{2}(1+\frac{t+\frac{3}{2}}{\frac{3}{2}})}=\frac{2}{3}\frac{1}{1-[-\frac{2}{3}(t-\frac{3}{2})]} $$ Reorganizing, and applying eq(1) $$\frac{2}{3}\frac{1}{1-[-\frac{2}{3}(t-\frac{3}{2})]}= \frac{2}{3}\sum_{k=0}^{\infty} [-\frac{2}{3}(t-\frac{3}{2})]^k $$
Putting $\frac{2}{3}$ inside the serie, and reorganizing $$\frac{2}{3}\frac{1}{1-[-\frac{2}{3}(t-\frac{3}{2})]}= \sum_{k=0}^{\infty} (-1)^k (\frac{2}{3})^{k+1}(t-\frac{3}{2})]^k $$
Now as we want $\ln(x)$, and we know that $\frac{2}{3}\frac{1}{1-[-\frac{2}{3}(t-\frac{3}{2})]}=\frac{1}{t}$, we will integrate as follow:
$$\ln(x)=\int_{1}^{x} \frac{1}{t} dt=\sum_{k=0}^{\infty} (-1)^k (\frac{2}{3})^{k+1}\int_{1}^{x}(t-\frac{3}{2})]^k dt$$
Obtaining: $$\ln(x)=\sum_{k=0}^{\infty} (-1)^k (\frac{2}{3})^{k+1} [\frac{(x-\frac{3}{2})^k}{k+1}-\frac{(\frac{1}{3})^k}{k+1} ]$$
Commom factor $\frac{1}{2}$ $$\ln(x)=\sum_{k=0}^{\infty} (-1)^k (\frac{2}{3})^{k+1} [\frac{(\frac{1}{2})^k(2x-3)^k-(\frac{1}{3})^k}{k+1}]$$
Reorganizing
$$\ln(x)=\sum_{k=0}^{\infty} (-1)^k (\frac{2}{3})^{k+1} (\frac{1}{2})^k[\frac{(2x-3)^k-(\frac{1}{3})^k}{k+1}]$$
But here I'm lacking of ideas, I do not how to continue. So please give a hand, a hint. Any help will be appreciated