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Evaluate-intdgral-xe-x-x-1-2-dx-please-help-

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Question Number 5893 by sanusihammed last updated on 04/Jun/16
Evaluate intdgral ((xe^x )/((x + 1)^2 )) dx please help.
$${Evaluate}\:{intdgral}\:\:\:\frac{{xe}^{{x}} }{\left({x}\:+\:\mathrm{1}\right)^{\mathrm{2}} }\:\:{dx}\: \\ $$$$ \\ $$$${please}\:{help}. \\ $$
Commented by FilupSmith last updated on 04/Jun/16
integrate by parts u=xe^x , dv=(1/((x+1)^2 ))dx ∴du=(x+1)e^x dx, ∴v=−(1/(x+1))dx ∫udv=uv−∫vdu ∴∫xe^x (1/((x+1)^2 ))dx=−((xe^x )/(x+1))+∫(1/(x+1))(x+1)e^x dx =−((xe^x )/(x+1))+e^x +c =e^x −((xe^x )/(x+1))+c =e^x (1−(x/(1+x)))+c =e^x ((((1+x)−x)/(1+x)))+c =e^x ((1/(1+x)))+c =(e^x /(1+x))+c, c=constant
$$\mathrm{integrate}\:\mathrm{by}\:\mathrm{parts} \\ $$$${u}={xe}^{{x}} ,\:\:\:\:\:\:{dv}=\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$$\therefore{du}=\left({x}+\mathrm{1}\right){e}^{{x}} {dx},\:\:\:\:\:\:\therefore{v}=−\frac{\mathrm{1}}{{x}+\mathrm{1}}{dx} \\ $$$$ \\ $$$$\int{udv}={uv}−\int{vdu} \\ $$$$ \\ $$$$\therefore\int{xe}^{{x}} \frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{dx}=−\frac{{xe}^{{x}} }{{x}+\mathrm{1}}+\int\frac{\mathrm{1}}{{x}+\mathrm{1}}\left({x}+\mathrm{1}\right){e}^{{x}} {dx} \\ $$$$=−\frac{{xe}^{{x}} }{{x}+\mathrm{1}}+{e}^{{x}} +{c} \\ $$$$={e}^{{x}} −\frac{{xe}^{{x}} }{{x}+\mathrm{1}}+{c} \\ $$$$={e}^{{x}} \left(\mathrm{1}−\frac{{x}}{\mathrm{1}+{x}}\right)+{c} \\ $$$$={e}^{{x}} \left(\frac{\left(\mathrm{1}+{x}\right)−{x}}{\mathrm{1}+{x}}\right)+{c} \\ $$$$={e}^{{x}} \left(\frac{\mathrm{1}}{\mathrm{1}+{x}}\right)+{c} \\ $$$$=\frac{{e}^{{x}} }{\mathrm{1}+{x}}+{c},\:\:\:\:{c}=\mathrm{constant} \\ $$
Commented by sanusihammed last updated on 04/Jun/16
Thanks a lot
$${Thanks}\:{a}\:{lot} \\ $$
Answered by Yozzii last updated on 04/Jun/16
∫((xe^x )/((x+1)^2 ))dx=∫((xe^x +e^x −e^x )/((x+1)^2 ))dx =∫(((x+1)e^x −1×e^x )/((x+1)^2 ))dx =∫(((x+1)(de^x /dx)−e^x ((d(x+1))/dx))/((x+1)^2 ))dx =∫(d/dx)((e^x /(1+x)))dx =(e^x /(1+x))+C
$$\int\frac{{xe}^{{x}} }{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{dx}=\int\frac{{xe}^{{x}} +{e}^{{x}} −{e}^{{x}} }{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$$=\int\frac{\left({x}+\mathrm{1}\right){e}^{{x}} −\mathrm{1}×{e}^{{x}} }{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$$=\int\frac{\left({x}+\mathrm{1}\right)\frac{{de}^{{x}} }{{dx}}−{e}^{{x}} \frac{{d}\left({x}+\mathrm{1}\right)}{{dx}}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$$=\int\frac{{d}}{{dx}}\left(\frac{{e}^{{x}} }{\mathrm{1}+{x}}\right){dx} \\ $$$$=\frac{{e}^{{x}} }{\mathrm{1}+{x}}+{C} \\ $$$$ \\ $$$$ \\ $$
Commented by FilupSmith last updated on 04/Jun/16
oh wow, although both methods are correct, I really like the simplicity of this method!
$$\mathrm{oh}\:\mathrm{wow},\:\mathrm{although}\:\mathrm{both}\:\mathrm{methods}\:\mathrm{are}\:\mathrm{correct}, \\ $$$$\mathrm{I}\:\mathrm{really}\:\mathrm{like}\:\mathrm{the}\:\mathrm{simplicity}\:\mathrm{of}\:\mathrm{this}\:\mathrm{method}! \\ $$
Commented by sanusihammed last updated on 04/Jun/16
Thanks so much
$${Thanks}\:{so}\:{much} \\ $$

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