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x-2-1-5-x-dx-




Question Number 89322 by I want to learn more last updated on 16/Apr/20
∫ (x^2 /(1 + 5^x )) dx
$$\int\:\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{1}\:+\:\mathrm{5}^{\mathrm{x}} }\:\mathrm{dx} \\ $$
Commented by mathmax by abdo last updated on 17/Apr/20
I =∫  (x^2 /(1+5^x ))dx ⇒I =∫  (x^2 /(1+e^(xln5) ))dx =_(e^(xln5) =t→x=((lnt)/(ln5))) (1/(ln5))  ∫  ((ln^2 (t))/(1+t))(dt/t)  ⇒I =(1/(ln5))∫ln^2 t((1/t)−(1/(1+t)))dt ⇒ln(5)×I =∫ ((ln^2 t)/t)dt−∫((ln^2 t)/(1+t))dt  by parts ∫ ((ln^2 t)/t)dt =ln^3 t−∫ ln(t)×((2lnt)/t) dt  =ln^3 t−2 ∫ ((ln^2 t)/t)dt ⇒∫((ln^2 t)/t)dt =(1/3)ln^3 (t) +c_0   ∫  ((ln^2 t)/(1+t))dt  =_(1+t =u)     ∫ ((ln^2 (u−1))/u) du  =_(by parts)    lnuln^2 (u−1)−∫ ln(u)×((2ln(u−1))/(u−1))du  =ln(u)ln^2 (u−1)−2 ∫ ((ln(u)ln(u−1))/(u−1))du ...be continued...
$${I}\:=\int\:\:\frac{{x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{5}^{{x}} }{dx}\:\Rightarrow{I}\:=\int\:\:\frac{{x}^{\mathrm{2}} }{\mathrm{1}+{e}^{{xln}\mathrm{5}} }{dx}\:=_{{e}^{{xln}\mathrm{5}} ={t}\rightarrow{x}=\frac{{lnt}}{{ln}\mathrm{5}}} \frac{\mathrm{1}}{{ln}\mathrm{5}}\:\:\int\:\:\frac{{ln}^{\mathrm{2}} \left({t}\right)}{\mathrm{1}+{t}}\frac{{dt}}{{t}} \\ $$$$\Rightarrow{I}\:=\frac{\mathrm{1}}{{ln}\mathrm{5}}\int{ln}^{\mathrm{2}} {t}\left(\frac{\mathrm{1}}{{t}}−\frac{\mathrm{1}}{\mathrm{1}+{t}}\right){dt}\:\Rightarrow{ln}\left(\mathrm{5}\right)×{I}\:=\int\:\frac{{ln}^{\mathrm{2}} {t}}{{t}}{dt}−\int\frac{{ln}^{\mathrm{2}} {t}}{\mathrm{1}+{t}}{dt} \\ $$$${by}\:{parts}\:\int\:\frac{{ln}^{\mathrm{2}} {t}}{{t}}{dt}\:={ln}^{\mathrm{3}} {t}−\int\:{ln}\left({t}\right)×\frac{\mathrm{2}{lnt}}{{t}}\:{dt} \\ $$$$={ln}^{\mathrm{3}} {t}−\mathrm{2}\:\int\:\frac{{ln}^{\mathrm{2}} {t}}{{t}}{dt}\:\Rightarrow\int\frac{{ln}^{\mathrm{2}} {t}}{{t}}{dt}\:=\frac{\mathrm{1}}{\mathrm{3}}{ln}^{\mathrm{3}} \left({t}\right)\:+{c}_{\mathrm{0}} \\ $$$$\int\:\:\frac{{ln}^{\mathrm{2}} {t}}{\mathrm{1}+{t}}{dt}\:\:=_{\mathrm{1}+{t}\:={u}} \:\:\:\:\int\:\frac{{ln}^{\mathrm{2}} \left({u}−\mathrm{1}\right)}{{u}}\:{du} \\ $$$$=_{{by}\:{parts}} \:\:\:{lnuln}^{\mathrm{2}} \left({u}−\mathrm{1}\right)−\int\:{ln}\left({u}\right)×\frac{\mathrm{2}{ln}\left({u}−\mathrm{1}\right)}{{u}−\mathrm{1}}{du} \\ $$$$={ln}\left({u}\right){ln}^{\mathrm{2}} \left({u}−\mathrm{1}\right)−\mathrm{2}\:\int\:\frac{{ln}\left({u}\right){ln}\left({u}−\mathrm{1}\right)}{{u}−\mathrm{1}}{du}\:…{be}\:{continued}… \\ $$
Commented by I want to learn more last updated on 17/Apr/20
Thanks sir, waiting
$$\mathrm{Thanks}\:\mathrm{sir},\:\mathrm{waiting} \\ $$

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