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Question Number 88089 by ar247 last updated on 08/Apr/20

∫(e^x /(e^2 −9))dx

$$\int\frac{{e}^{{x}} }{{e}^{\mathrm{2}} −\mathrm{9}}{dx} \\ $$

Commented by ar247 last updated on 08/Apr/20

help

$${help} \\ $$

Answered by Rio Michael last updated on 08/Apr/20

∫(e^x /(e^2 −9))dx =(1/(e^2 −9)) ∫e^x dx = (e^x /(e^2 −9)) + k

$$\int\frac{{e}^{{x}} }{{e}^{\mathrm{2}} −\mathrm{9}}{dx}\:=\frac{\mathrm{1}}{{e}^{\mathrm{2}} −\mathrm{9}}\:\int{e}^{{x}} {dx}\:=\:\frac{{e}^{{x}} }{{e}^{\mathrm{2}} −\mathrm{9}}\:+\:{k}\: \\ $$

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