3-ln-x-2-dx-

Facebook Tweet Pin

Question Number 91753 by Rio Michael last updated on 02/May/20

$$\int\mathrm{3}\:\left(\mathrm{ln}\:{x}\right)^{\mathrm{2}} \:{dx}\:=\:?? \\ $$

Commented by peter frank last updated on 02/May/20

$${by}\:{part} \\ $$

Commented by mathmax by abdo last updated on 03/May/20

I =3∫ (lnx)^2 dx changement ln(x)=t give I =3 ∫ t^2 e^t dt =_(by psrts) 3{ t^2 e^t −∫ 2t e^t } =3{ t^2 e^t −2( te^t −∫e^t dt)} =3{ t^2 e^t −2te^t +2 e^t ) +c =(3t^2 −6t +6)e^t +c =(3(lnx)^2 −6ln(x)+6)x +c =3x(lnx)^2 −6xln(x)+6x +c

$${I}\:=\mathrm{3}\int\:\left({lnx}\right)^{\mathrm{2}} \:{dx}\:{changement}\:{ln}\left({x}\right)={t}\:{give} \\ $$$${I}\:=\mathrm{3}\:\int\:{t}^{\mathrm{2}} {e}^{{t}} \:{dt}\:\:=_{{by}\:{psrts}} \:\:\:\mathrm{3}\left\{\:\:{t}^{\mathrm{2}} \:{e}^{{t}} \:−\int\:\mathrm{2}{t}\:{e}^{{t}} \right\} \\ $$$$=\mathrm{3}\left\{\:{t}^{\mathrm{2}} \:{e}^{{t}} −\mathrm{2}\left(\:\:{te}^{{t}} −\int{e}^{{t}} \:{dt}\right)\right\} \\ $$$$=\mathrm{3}\left\{\:{t}^{\mathrm{2}} \:{e}^{{t}} −\mathrm{2}{te}^{{t}} \:+\mathrm{2}\:{e}^{{t}} \right)\:+{c} \\ $$$$=\left(\mathrm{3}{t}^{\mathrm{2}} −\mathrm{6}{t}\:+\mathrm{6}\right){e}^{{t}} \:+{c} \\ $$$$=\left(\mathrm{3}\left({lnx}\right)^{\mathrm{2}} −\mathrm{6}{ln}\left({x}\right)+\mathrm{6}\right){x}\:+{c} \\ $$$$=\mathrm{3}{x}\left({lnx}\right)^{\mathrm{2}} −\mathrm{6}{xln}\left({x}\right)+\mathrm{6}{x}\:+{c} \\ $$

Commented by Tony Lin last updated on 03/May/20

3∫(lnx)^n dx let t=lnx,(dt/dx)=(1/x)=(1/e^t ) 3∫e^t t^n dt =3e^t Σ_(k=0) ^n (t^n )^((k)) (−1)^k =3xΣ_(k=0) ^n (lnx)^k P_k ^n (−1)^k ,P_k ^n =C_k ^n ×k!

$$\mathrm{3}\int\left({lnx}\right)^{{n}} {dx} \\ $$$${let}\:{t}={lnx},\frac{{dt}}{{dx}}=\frac{\mathrm{1}}{{x}}=\frac{\mathrm{1}}{{e}^{{t}} } \\ $$$$\mathrm{3}\int{e}^{{t}} {t}^{{n}} {dt} \\ $$$$=\mathrm{3}{e}^{{t}} \underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left({t}^{{n}} \right)^{\left({k}\right)} \left(−\mathrm{1}\right)^{{k}} \\ $$$$=\mathrm{3}{x}\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left({lnx}\right)^{{k}} {P}_{{k}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} ,{P}_{{k}} ^{{n}} ={C}_{{k}} ^{{n}} ×{k}! \\ $$

Answered by M±th+et+s last updated on 02/May/20

I=∫ln^2 (x)dx let u=ln^2 (x) dv=dx du=((2ln(x))/x) dx v=x I=∫udv ∫udv=uv−∫vdu =xln^2 (x)−∫x ((2ln(x))/x) dx =xln^2 (x)−2∫ln(x)dx =xln^2 (x)−2(xln(x)−x)+c ∫3ln^2 (x)dx=3I

$${I}=\int{ln}^{\mathrm{2}} \left({x}\right){dx} \\ $$$${let}\:{u}={ln}^{\mathrm{2}} \left({x}\right)\:\:\:\:\:\:{dv}={dx} \\ $$$$\:\:\:\:\:\:\:{du}=\frac{\mathrm{2}{ln}\left({x}\right)}{{x}}\:{dx}\:\:\:{v}={x} \\ $$$${I}=\int{udv} \\ $$$$\int{udv}={uv}−\int{vdu} \\ $$$$={xln}^{\mathrm{2}} \left({x}\right)−\int{x}\:\frac{\mathrm{2}{ln}\left({x}\right)}{{x}}\:{dx} \\ $$$$={xln}^{\mathrm{2}} \left({x}\right)−\mathrm{2}\int{ln}\left({x}\right){dx} \\ $$$$={xln}^{\mathrm{2}} \left({x}\right)−\mathrm{2}\left({xln}\left({x}\right)−{x}\right)+{c} \\ $$$$\int\mathrm{3}{ln}^{\mathrm{2}} \left({x}\right){dx}=\mathrm{3}{I} \\ $$$$ \\ $$

Commented by Rio Michael last updated on 02/May/20

$$\mathrm{thanks}\:\mathrm{sir} \\ $$

Commented by M±th+et+s last updated on 02/May/20

$${you}\:{are}\:{welcome} \\ $$

Facebook Tweet Pin

Leave a Reply Cancel reply