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0-3-e-2x-dx-

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Question Number 88104 by ar247 last updated on 08/Apr/20
∫_0 ^(+∞) (√(3+e^(−2x) ))dx
∫0+∞3+e−2xdx
Commented by ar247 last updated on 08/Apr/20
with stap please
withstapplease
Commented by mathmax by abdo last updated on 08/Apr/20
let take a try I =∫_0 ^∞ (√(3+e^(−2x) ))dx changement (√(3+e^(−2x) ))=t give 3+e^(−2x) =t^2 ⇒e^(−2x) =t^2 −3 ⇒−2x =ln(t^2 −3) ⇒x =−(1/2)ln(t^2 −3) ⇒ I =∫_2 ^(√3) t(−(1/2))×((2t)/(t^2 −3))dt = ∫_(√3) ^2 (t^2 /(t^2 −3))dt =∫_(√3) ^2 ((t^2 −3+3)/(t^2 −3))dt =2−(√3)+3∫_(√3) ^2 (dt/(t^2 −3)) =2−(√3)+(3/(2(√3))) ∫_(√3) ^2 ((1/(t−(√3)))−(1/(t+(√3))))dt =2−(√3)+((√3)/2) [ln∣((t−(√3))/(t+(√3)))∣]_(√3) ^2 +C we have lim_(t→(√3)) ln∣((t−(√3))/(t+(√3)))∣=−∞ so this integral is dvergent...!
lettakeatryI=∫0∞3+e−2xdxchangement3+e−2x=tgive3+e−2x=t2⇒e−2x=t2−3⇒−2x=ln(t2−3)⇒x=−12ln(t2−3)⇒I=∫23t(−12)×2tt2−3dt=∫32t2t2−3dt=∫32t2−3+3t2−3dt=2−3+3∫32dtt2−3=2−3+323∫32(1t−3−1t+3)dt=2−3+32[ln∣t−3t+3∣]32+Cwehavelimt→3ln∣t−3t+3∣=−∞sothisintegralisdvergent…!
Answered by Joel578 last updated on 08/Apr/20
If x→+∞, then 3 + e^(−2x) → 3, hence ∫_0 ^( +∞) (√(3 + e^(−2x) )) dx = +∞
Ifx→+∞,then3+e−2x→3,hence∫0+∞3+e−2xdx=+∞

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