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Question Number 135824 by Khakie last updated on 16/Mar/21

$${\int }_0^a\frac{x^4{e}^x}{{(e^x-1)}^2}dt$$$$$$$$pleasesolvethis...$$

Commented by mathmax by abdo last updated on 16/Mar/21

$$\mathrm{i}\mathrm{think}\mathrm{the}\mathrm{Q}\mathrm{is}{\int }_0^{\mathrm{a}}\frac{{\mathrm{x}}^4{\mathrm{e}}^{\mathrm{x}}}{{({\mathrm{e}}^{\mathrm{x}}-1)}^2}\mathrm{dx}..$$

Answered by Ñï= last updated on 16/Mar/21

$${\int }_0^a\frac{x^4{e}^x}{{(e^x-1)}^2}dt=\frac{x^4{e}^x}{{(e^z-1)}^2}t{\mid }_0^a=\frac{x^4{e}^x}{{(e^x-1)}^2}a$$

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