Taylor series expansion of $\ln(5 + x)$. If I start with the function $1/(5+x)$ I can find its Taylor series about 0 by using the familiar geometric series $1/(1 + x)$.
\begin{align}
\frac{1}{5 + x} &= \frac{1}{5 (1 + x/5)} \\
&= \frac15 \cdot \frac{1}{1 + (1/5) x} \\
&= \frac15 - \frac{x}{25} + \frac{x^2}{125} - \frac{x^3}{625} + \frac{x^4}{3125} - \frac{x^5}{15625} + \dots
\end{align}
If I take the antiderivative of both sides I get:
$$ \ln(5 + x) = \frac{x}{5} - \frac{x^2}{50} + \frac{x^3}{375} - \frac{x^4}{2500} + \frac{x^5}{15625} - \dots$$
So in particular I have $\ln(5) = 0$.

First the correct answer:

If you do the transformation without integration you get
$$ \ln(5 + x) = \ln \left(5 \left(1 + \frac15 x\right) \right) = \ln 5 + \ln \left(1 + \frac15 x\right). $$
Then using the series for $\ln(1 + x)$ you get
$$ \ln(5 + x) = \ln 5 + \sum_{n = 1}^\infty (-1)^{n + 1} \frac{x^n}{n \cdot 5^n}. $$

This is the same as your answer except for the $\ln 5$ in front.
It comes down to the fact that there is no difference between the derivatives of $\ln(5 + x)$ and $\ln(5 + x) - \ln 5$. When you take the integral you have to account for the constant term seperately.
