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calculate-2-5-e-x-1-x-1-dx-




Question Number 35427 by abdo.msup.com last updated on 18/May/18
calculate   ∫_2 ^5    (e^(√(x+1)) /( (√(x+1))))dx
$${calculate}\:\:\:\int_{\mathrm{2}} ^{\mathrm{5}} \:\:\:\frac{{e}^{\sqrt{{x}+\mathrm{1}}} }{\:\sqrt{{x}+\mathrm{1}}}{dx} \\ $$
Commented by prof Abdo imad last updated on 19/May/18
changement (√(x+1))=t give x=t^2 −1  I  = ∫_(√3) ^(√6)     (e^t /t) 2 tdt = 2 ∫_(√3) ^(√6)   e^t dt  = 2{  e^(√6)   −e^(√3) } .
$${changement}\:\sqrt{{x}+\mathrm{1}}={t}\:{give}\:{x}={t}^{\mathrm{2}} −\mathrm{1} \\ $$$${I}\:\:=\:\int_{\sqrt{\mathrm{3}}} ^{\sqrt{\mathrm{6}}} \:\:\:\:\frac{{e}^{{t}} }{{t}}\:\mathrm{2}\:{tdt}\:=\:\mathrm{2}\:\int_{\sqrt{\mathrm{3}}} ^{\sqrt{\mathrm{6}}} \:\:{e}^{{t}} {dt} \\ $$$$=\:\mathrm{2}\left\{\:\:{e}^{\sqrt{\mathrm{6}}} \:\:−{e}^{\sqrt{\mathrm{3}}} \right\}\:. \\ $$
Answered by ajfour last updated on 19/May/18
I=2∫_(√3) ^(√6) e^t dt =2(e^(√6) −e^(√3) ) .
$${I}=\mathrm{2}\int_{\sqrt{\mathrm{3}}} ^{\sqrt{\mathrm{6}}} {e}^{{t}} {dt}\:=\mathrm{2}\left({e}^{\sqrt{\mathrm{6}}} −{e}^{\sqrt{\mathrm{3}}} \right)\:. \\ $$

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