Let $\mathrm{I}_{\mathrm{n}}=\int_{1}^{\mathrm{c}} \mathrm{x}^{19}(\log |\mathrm{x}|)^{\mathrm{n}} \mathrm{dx}$, where $\mathrm{n} \in \mathrm{N}$. If
(20) $\mathrm{I}_{10}=\alpha \mathrm{I}_{9}+\beta \mathrm{I}_{8}$, for natural numbers $\alpha$ and $\beta$, then $\alpha-\beta$ equal to
$\mathrm{I}_{\mathrm{n}}=\int_{1}^{\mathrm{e}} \mathrm{x}^{19}(\log |\mathrm{x}|)^{\mathrm{n}} \mathrm{d} \mathrm{x}$
$I_{n}=\left|(\log |x|)^{19} \frac{x^{20}}{20}\right|_{1}^{e}-\int n(\log |x|)^{n-1} \cdot \frac{1}{x} \cdot \frac{x^{20}}{20} d x$
$20 \mathrm{I}_{\mathrm{n}}=\mathrm{e}^{20}-\mathrm{nI}_{\mathrm{n}-1}$
$\therefore 20 \mathrm{I}_{10}=\mathrm{e}^{20}-10 \mathrm{I}_{9}$
$20 \mathrm{I}_{9}=\mathrm{e}^{20}-9 \mathrm{I}_{8}$
$\Rightarrow 20 \mathrm{I}_{10}=10 \mathrm{I}_{9}+9 \mathrm{I}_{8}$
$\alpha=10, \beta=9$
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