Menu Close

e-x-e-2-9-dx-




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

Leave a Reply

Your email address will not be published. Required fields are marked *