## Calculus (3rd Edition)

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