$\text{ If }{\mathrm{I}}_{\mathrm{n}}={{\int }_1^e}^{ }(\ln \mathrm{x}{)}^{\mathrm{n}}\mathrm{dx}\text{ then }{\mathrm{I}}_8+8{\mathrm{I}}_7=$

$$\begin{gathered} {\mathrm{I}}_{\mathrm{n}}={{\int }_1^e}^{ }(\ln \mathrm{x}{)}^{\mathrm{n}}\mathrm{dx} \\ sub n=8 on both sides \\ {\mathrm{I}}_8={\int }_1^{\mathrm{e}}(\mathrm{lnx}{)}^8 \mathrm{dx} \\ {\mathrm{I}}_8={{\int }_1^e}^{ }(\mathrm{lnx}{)}^8·1 \mathrm{dx} \\ By applying by parts formula \\ {\left.{\mathrm{I}}_8=(\mathrm{logx}{)}^8\mathrm{x}\right]}_1^{\mathrm{e}}-\int 8(\log \mathrm{x}{)}^7\frac{1}{x}\left(\mathrm{x}\right)\mathrm{dx} \\ =\left[{\left(\log e\right)}^{8}e-0\right]-8\int {\left(\log x\right)}^{7}dx \\ \Rightarrow I_8=\mathrm{e}-8{\mathrm{I}}_7 \\ \Rightarrow {\mathrm{I}}_8+8{\mathrm{I}}_7=\mathrm{e} \end{gathered}$$