Suppose, I want to find a function such that its Taylor series expansion is $$f(x) = \sum_{n=0}^{\infty}\frac{x^{n+1}}{(n+1)a^n}$$
I could start with $$\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n$$
Integrate it, substitute $x\rightarrow \frac{x}{a}$, multiply by $a$ and get
$$F(x) = -\ln|x-1| = \sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}$$
$$a F\left(\frac{x}{a}\right) = -a \ln\left|\frac{x}{a}-1\right| = \sum_{n=0}^{\infty}\frac{x^{n+1}}{(n+1)a^n}$$
On the other hand, I could start with subtituting $x \rightarrow \frac{x}{a}$ before integration to get
$$\frac{a}{a-x} = \sum_{n=0}^{\infty}\frac{x^n}{a^n}$$
and then integrate it to get $$-a\ln|x-a| = \sum_{n=0}^{\infty}\frac{x^{n+1}}{(n+1)a^n}$$
As you can see, arguments of $\ln$ are not equal. Where did it go wrong?
When you integrate, you should include a constant of integration. What you see here is that when integrating the functions, you get different constants of integration. This is why your answers differ by only a constant, namely $a\ln a$ (you can see this by use of $\log$ rules).
If you take care with the limits or boundary conditions in the integration step, then the answers will agree exactly.