Solutions

a) We just need to think of differentiation in reverse. One function $g$ that satisfies $g'=f$ is

$$g(x)=\frac{a_n}{n+1}x^{n+1}+\frac{a_{n-1}}{n}x^n+\cdots+a_0x.$$

Since the derivative of any constant is $0$, any function $h(x)=g(x)+c$ for any $c$ also satisfies $h'=f$.

b) We have

$$g(x)=-\frac{b_2}{x}-\frac{b_3}{3x^2}-\cdots-\frac{b_m}{mx^{m-1}}.$$

c) We have that

$$f'(x)=na_nx^{n-1}+(n-1)a_{n-1}x^{n-2}+\cdots+a_1-\frac{b_1}{x^2}-\cdots-\frac{mb_m}{x^{m+1}}.$$

Hence, there are no coefficients whereby $f'(x)=\frac{1}{x}$.