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0-n-10-e-x-e-x-1-e-x-8-dx-




Question Number 122980 by liberty last updated on 21/Nov/20
 ∫_0 ^(ℓn 10)  ((e^x  (√(e^x −1)))/(e^x +8)) dx ?
$$\:\int_{\mathrm{0}} ^{\ell{n}\:\mathrm{10}} \:\frac{{e}^{{x}} \:\sqrt{{e}^{{x}} −\mathrm{1}}}{{e}^{{x}} +\mathrm{8}}\:{dx}\:? \\ $$
Answered by bemath last updated on 21/Nov/20
Answered by mathmax by abdo last updated on 21/Nov/20
I =∫_0 ^(ln(10))  ((e^x (√(e^x −1)))/(e^x  +8))dx  changement (√(e^x −1))=t give  e^x −1 =t^2  ⇒e^x  =t^2  +1 ⇒x=ln(t^2  +1) ⇒  I =∫_0 ^3    (((t^2 +1)t)/(t^2  +1+8))×((2tdt)/(t^2  +1)) =∫_0 ^3   ((2t^2 )/(t^2  +9))dt  =2∫_0 ^3  ((t^2 +9−9)/(t^2  +9))dt =2∫_0 ^3 dt −18∫_0 ^3  (dt/(t^2  +9))(→t=3u)  =6−18  ∫_0 ^1  ((3du)/(9(1+u^2 ))) =6−6 ∫_0 ^1  (du/(1+u^2 )) =6−6[arctanu]_0 ^1   =6−6.(π/4) =6−((3π)/2)
$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\mathrm{ln}\left(\mathrm{10}\right)} \:\frac{\mathrm{e}^{\mathrm{x}} \sqrt{\mathrm{e}^{\mathrm{x}} −\mathrm{1}}}{\mathrm{e}^{\mathrm{x}} \:+\mathrm{8}}\mathrm{dx}\:\:\mathrm{changement}\:\sqrt{\mathrm{e}^{\mathrm{x}} −\mathrm{1}}=\mathrm{t}\:\mathrm{give} \\ $$$$\mathrm{e}^{\mathrm{x}} −\mathrm{1}\:=\mathrm{t}^{\mathrm{2}} \:\Rightarrow\mathrm{e}^{\mathrm{x}} \:=\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\:\Rightarrow\mathrm{x}=\mathrm{ln}\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)\:\Rightarrow \\ $$$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\mathrm{3}} \:\:\:\frac{\left(\mathrm{t}^{\mathrm{2}} +\mathrm{1}\right)\mathrm{t}}{\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}+\mathrm{8}}×\frac{\mathrm{2tdt}}{\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}}\:=\int_{\mathrm{0}} ^{\mathrm{3}} \:\:\frac{\mathrm{2t}^{\mathrm{2}} }{\mathrm{t}^{\mathrm{2}} \:+\mathrm{9}}\mathrm{dt} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{3}} \:\frac{\mathrm{t}^{\mathrm{2}} +\mathrm{9}−\mathrm{9}}{\mathrm{t}^{\mathrm{2}} \:+\mathrm{9}}\mathrm{dt}\:=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{3}} \mathrm{dt}\:−\mathrm{18}\int_{\mathrm{0}} ^{\mathrm{3}} \:\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} \:+\mathrm{9}}\left(\rightarrow\mathrm{t}=\mathrm{3u}\right) \\ $$$$=\mathrm{6}−\mathrm{18}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{3du}}{\mathrm{9}\left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)}\:=\mathrm{6}−\mathrm{6}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{du}}{\mathrm{1}+\mathrm{u}^{\mathrm{2}} }\:=\mathrm{6}−\mathrm{6}\left[\mathrm{arctanu}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\mathrm{6}−\mathrm{6}.\frac{\pi}{\mathrm{4}}\:=\mathrm{6}−\frac{\mathrm{3}\pi}{\mathrm{2}} \\ $$

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