#### Answer

$$f (x)=4+ \sum_{n=1}^{\infty} \frac{(x-3)^{n+1}}{n(n+1)}$$

#### Work Step by Step

Given $$f^{\prime}(x)=\sum_{n=1}^{\infty} \frac{(x-3)^{n}}{n}$$
By integration, we get
$$f (x)=C+ \sum_{n=1}^{\infty} \frac{(x-3)^{n+1}}{n(n+1)}$$
Since $f(3)=4 $, then $C=4$, hence
$$f (x)=4+ \sum_{n=1}^{\infty} \frac{(x-3)^{n+1}}{n(n+1)}$$