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Question Number 122980 by liberty last updated on 21/Nov/20

$${\int }_0^{\ell n10}\frac{e^x\sqrt{e^x-1}}{e^x+8}dx?$$

Answered by bemath last updated on 21/Nov/20

Answered by mathmax by abdo last updated on 21/Nov/20

$$\mathrm{I}={\int }_0^{\ln \left(10\right)}\frac{{\mathrm{e}}^{\mathrm{x}}\sqrt{{\mathrm{e}}^{\mathrm{x}}-1}}{{\mathrm{e}}^{\mathrm{x}}+8}\mathrm{dx}\mathrm{changement}\sqrt{{\mathrm{e}}^{\mathrm{x}}-1}=\mathrm{t}\mathrm{give}$$$${\mathrm{e}}^{\mathrm{x}}-1={\mathrm{t}}^2\Rightarrow {\mathrm{e}}^{\mathrm{x}}={\mathrm{t}}^2+1\Rightarrow \mathrm{x}=\ln ({\mathrm{t}}^2+1)\Rightarrow$$$$\mathrm{I}={\int }_0^3\frac{({\mathrm{t}}^2+1)\mathrm{t}}{{\mathrm{t}}^2+1+8}\times \frac{2\mathrm{tdt}}{{\mathrm{t}}^2+1}={\int }_0^3\frac{{2\mathrm{t}}^2}{{\mathrm{t}}^2+9}\mathrm{dt}$$$$=2{\int }_0^3\frac{{\mathrm{t}}^2+9-9}{{\mathrm{t}}^2+9}\mathrm{dt}=2{\int }_0^3\mathrm{dt}-18{\int }_0^3\frac{\mathrm{dt}}{{\mathrm{t}}^2+9}(\to \mathrm{t}=3\mathrm{u})$$$$=6-18{\int }_0^1\frac{3\mathrm{du}}{9(1+{\mathrm{u}}^2)}=6-6{\int }_0^1\frac{\mathrm{du}}{1+{\mathrm{u}}^2}=6-6{\left[\mathrm{arctanu}\right]}_0^1$$$$=6-6.\frac{\pi }{4}=6-\frac{3\pi }{2}$$

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