Let's define
\begin{equation*} I_0 := \log\frac{6}{5} \end{equation*} and for $k = 1, 2, \ldots, n$ \begin{equation*} I_k := \frac{1}{k} - 5 I_{k-1}. \end{equation*}
How the value $I_n$ is linked with the value of $$\int_0^1\frac{x^n}{x+5} \mathrm{d}x \ ?$$
If we set $$ I_n = \int_{0}^{1}\frac{x^n}{x+5}\,dx \tag{1}$$ we clearly have $I_0=\log\frac{6}{5}$ and $$ I_n+ 5I_{n-1} = \int_{0}^{1}\frac{x^n+5 x^{n-1}}{x+5}\,dx = \int_{0}^{1}x^{n-1}\,dx = \frac{1}{n}.\tag{2} $$