Question Number 216372 by glory86 last updated on 06/Feb/25
$$\int\frac{{xe}^{{x}} }{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$
Answered by MrGaster last updated on 06/Feb/25
$$\mathrm{Let}\:{u}={x}+\mathrm{1}\Rightarrow{du}={dx},{x}={u}−\mathrm{1} \\ $$$$\int\frac{\left({u}−\mathrm{1}\right){e}^{{u}−\mathrm{1}} }{{u}^{\mathrm{2}} }{du}=\int\frac{{ue}^{{u}−\mathrm{1}} }{{u}^{\mathrm{2}} }{du}−\int\frac{{e}^{{u}−\mathrm{1}} }{{u}^{\mathrm{2}} }{du} \\ $$$$=\int\frac{{e}^{{u}−\mathrm{1}} }{{u}}{du}−\int\frac{{e}^{{u}−\mathrm{1}} }{{u}^{\mathrm{2}} }{du} \\ $$$$=\int\frac{{e}^{{u}} }{{ue}}{du}−\int\frac{{e}^{{u}} }{{u}^{\mathrm{2}} {e}}{du} \\ $$$$=\frac{\mathrm{1}}{{e}}\left(\int\frac{{e}^{{u}} }{{u}}{du}−\int\frac{{e}^{{u}} }{{u}^{\mathrm{2}} }\right){du} \\ $$$$=\frac{\mathrm{1}}{{e}}\left(\mathrm{Ei}\left({u}\right)+\frac{{e}^{{u}} }{{u}}\right)+{C} \\ $$$$=\frac{{e}^{{u}−\mathrm{1}} }{{u}}+{C}=\frac{{e}^{{x}} }{{x}+\mathrm{1}}+{C} \\ $$
Answered by som(math1967) last updated on 06/Feb/25
$$\:\int\frac{{xe}^{{x}} +{e}^{{x}} −{e}^{{x}} }{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$$=\int\frac{{e}^{{x}} \left({x}+\mathrm{1}\right)}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{dx}−\int\frac{{e}^{{x}} }{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$$=\int\frac{{e}^{{x}} {dx}}{\left({x}+\mathrm{1}\right)}\:−\int\frac{{e}^{{x}} }{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$$=\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)}\int{e}^{{x}} {dx}−\int\left\{\frac{{d}}{{dx}}×\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)}\int{e}^{{x}} {dx}\right\}{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\int\frac{{e}^{{x}} }{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$$=\frac{{e}^{{x}} }{\left({x}+\mathrm{1}\right)}\:+\int\frac{{e}^{{x}} }{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{dx}\:−\int\frac{{e}^{{x}} }{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$$=\frac{{e}^{{x}} }{\left({x}+\mathrm{1}\right)}\:+{C} \\ $$