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e-x-1-e-x-dx-




Question Number 6699 by jose14918645@gmail.com last updated on 14/Jul/16
∫(e^x +1)e^x dx
$$\int\left({e}^{{x}} +\mathrm{1}\right){e}^{{x}} {dx} \\ $$
Answered by FilupSmith last updated on 14/Jul/16
=∫e^(2x) +e^x dx  =(1/2)e^(2x) +e^x +c
$$=\int{e}^{\mathrm{2}{x}} +{e}^{{x}} {dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{e}^{\mathrm{2}{x}} +{e}^{{x}} +{c} \\ $$
Answered by FilupSmith last updated on 15/Jul/16
u=e^x   du=e^x dx=udx  =∫(u+1)udx  =∫(u+1)du  =(1/2)u^2 +u+c  =(1/2)e^(2x) +e^x +c
$${u}={e}^{{x}} \\ $$$${du}={e}^{{x}} {dx}={udx} \\ $$$$=\int\left({u}+\mathrm{1}\right){udx} \\ $$$$=\int\left({u}+\mathrm{1}\right){du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{u}^{\mathrm{2}} +{u}+{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{e}^{\mathrm{2}{x}} +{e}^{{x}} +{c} \\ $$

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