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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 $$\int_0^1\frac{x^n}{x+5} \mathrm{d}x \ ?$$

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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 $$ I_n+ 5I_{n-1} = \int_{0}^{1}\frac{x^n+5 x^{n-1}}{x+5}\,dx = \int_{0}^{1}x^{n-1}\,dx = \frac{1}{n}.\tag{2} $$

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$I_n+5I_{n-1}=\int_{0}^{1}\frac{x^n}{x+5}dx+5\int_{0}^{1}\frac{x^{n-1}}{x+5}dx=\int_{0}^{1}(\frac{x^n}{x+5}+5\frac{x^{n-1}}{x+5})dx$

$I_n+5I_{n-1}=\int_{0}^{1}x^{n-1}dx=\frac{x^n}{n}|_{0}^{1}=\frac{1}{n}$

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