I was trying different approaches to a problem involving power series. I'm requesting to represent $\ln(x)$ as a power series with poweres of $(x-4)$. I wanted to see if I could use the expantion
$$\frac{1}{1-x}=\sum_{n=0}^\infty x^n$$
Let $u=x-4$, so that $f(x)=\ln(x)=\ln(u+4)=\ln(4(1+\frac{u}{4})=\ln(4)+\ln(1+\frac{u}{4})$
Then
$$\frac{d}{dx} \ln(x)= \frac{d}{du}(\ln(4)+\ln(1+\frac{u}{4}))=0+\frac{d}{du}\ln(1+\frac{u}{4})$$
$$=\frac{1}{1-(-\frac{u}{4})}\frac{1}{4}=\frac{1}{4}\sum_{n=0}^\infty(-\frac{u}{4})^n=\sum_{n=0}^\infty(-1)^n(\frac{x-4}{4})^n\frac{1}{4}=\sum_{n=0}^\infty(-1)^n\frac{(x-4)^n}{4^{n+1}}$$
Because $\frac{d}{dx} \ln x= \sum_{n=0}^\infty(-1)^n\frac{(x-4)^n}{4^{n+1}}$, then
$$\ln x= \int\sum_{n=0}^\infty(-1)^n\frac{(x-4)^n}{4^{n+1}} dx=\sum_{n=0}^\infty\int((-1)^n\frac{(x-4)^n}{4^{n+1}} dx)=\sum_{n=0}^\infty(-1)^n\frac{(x-4)^{n+1}}{4^{n+1}(n+1)}$$
$$=C+\sum_{n=1}^\infty(-1)^{n-1}\frac{(x-4)^n}{n4^{n}}$$
That is almost correct. However, that $C$ should not be part of the final answer ($\log x$ is represented by one power series centered at $4$). In fact, $C=\log 4$.
Also, you should no have written that $\displaystyle\log(x)=\int\sum_{n=0}^\infty(-1)^n\frac{(x-4)^n}{4^{n+1}}$; in fact,$$\displaystyle\log(x)=\log(4)+\int_4^x\sum_{n=0}^\infty(-1)^n\frac{(x-4)^n}{4^{n+1}}.$$