Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 14243 by tawa tawa last updated on 30/May/17

∫ [x (lnx)^2 ] dx

$$\int\:\left[\mathrm{x}\:\left(\mathrm{lnx}\right)^{\mathrm{2}} \right]\:\mathrm{dx} \\ $$

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 30/May/17

I=∫x^2 ln^2 xdx  t=lnx⇒dx=xdt=e^t dt  I=∫e^(2t) .t^2 e^t dt=∫t^2 e^(3t) dt=  =t^2 (1/3)e^(3t) −∫2t(1/3)e^(3t) dt=  =((t^2 e^(3t) )/3)−(2/3)t.(1/3)e^(3t) +∫(1/3)e^(3t) dt=  =(1/9)(3t^2 e^(3t) −2te^(3t) +e^(3t) )+C=  =(x^3 /9).(3ln^2 x−2lnx+1)+C  .■

$${I}=\int{x}^{\mathrm{2}} {ln}^{\mathrm{2}} {xdx} \\ $$$${t}={lnx}\Rightarrow{dx}={xdt}={e}^{{t}} {dt} \\ $$$${I}=\int{e}^{\mathrm{2}{t}} .{t}^{\mathrm{2}} {e}^{{t}} {dt}=\int{t}^{\mathrm{2}} {e}^{\mathrm{3}{t}} {dt}= \\ $$$$={t}^{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{3}}{e}^{\mathrm{3}{t}} −\int\mathrm{2}{t}\frac{\mathrm{1}}{\mathrm{3}}{e}^{\mathrm{3}{t}} {dt}= \\ $$$$=\frac{{t}^{\mathrm{2}} {e}^{\mathrm{3}{t}} }{\mathrm{3}}−\frac{\mathrm{2}}{\mathrm{3}}{t}.\frac{\mathrm{1}}{\mathrm{3}}{e}^{\mathrm{3}{t}} +\int\frac{\mathrm{1}}{\mathrm{3}}{e}^{\mathrm{3}{t}} {dt}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{9}}\left(\mathrm{3}{t}^{\mathrm{2}} {e}^{\mathrm{3}{t}} −\mathrm{2}{te}^{\mathrm{3}{t}} +{e}^{\mathrm{3}{t}} \right)+{C}= \\ $$$$=\frac{{x}^{\mathrm{3}} }{\mathrm{9}}.\left(\mathrm{3}{ln}^{\mathrm{2}} {x}−\mathrm{2}{lnx}+\mathrm{1}\right)+{C}\:\:.\blacksquare \\ $$

Commented by tawa tawa last updated on 30/May/17

sir, i think the square is affecting only  lnx  as in   [x (lnx)^2 ] dx

$$\mathrm{sir},\:\mathrm{i}\:\mathrm{think}\:\mathrm{the}\:\mathrm{square}\:\mathrm{is}\:\mathrm{affecting}\:\mathrm{only}\:\:\mathrm{lnx} \\ $$$$\mathrm{as}\:\mathrm{in}\:\:\:\left[\mathrm{x}\:\left(\mathrm{lnx}\right)^{\mathrm{2}} \right]\:\mathrm{dx} \\ $$

Commented by tawa tawa last updated on 30/May/17

Not    [x ln(x)]^2      it is   [x (lnx)^2 ] dx

$$\mathrm{Not}\:\:\:\:\left[\mathrm{x}\:\mathrm{ln}\left(\mathrm{x}\right)\right]^{\mathrm{2}} \:\:\:\:\:\mathrm{it}\:\mathrm{is}\:\:\:\left[\mathrm{x}\:\left(\mathrm{lnx}\right)^{\mathrm{2}} \right]\:\mathrm{dx} \\ $$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 30/May/17

you are right.your answer is given  below by dear mr Ajfour.excuse me.

$${you}\:{are}\:{right}.{your}\:{answer}\:{is}\:{given} \\ $$$${below}\:{by}\:{dear}\:{mr}\:{Ajfour}.{excuse}\:{me}. \\ $$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 30/May/17

additionaly now you have answer of  another similar integral .▶◂

$${additionaly}\:{now}\:{you}\:{have}\:{answer}\:{of} \\ $$$${another}\:{similar}\:{integral}\:.\blacktriangleright\blacktriangleleft \\ $$

Commented by chux last updated on 30/May/17

thanks sir...... i really love this

$$\mathrm{thanks}\:\mathrm{sir}......\:\mathrm{i}\:\mathrm{really}\:\mathrm{love}\:\mathrm{this} \\ $$

Commented by tawa tawa last updated on 30/May/17

I really appreciate . God bless you sir

$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Answered by ajfour last updated on 30/May/17

let ln x=t   ⇒ x=e^t   and   dx=e^t dt   I=∫[x(ln x)^2 ]dx     = ∫e^t (t^2 )(e^t dt)=∫t^2 e^(2t) dt     = (t^2 /2)e^(2t) −∫((e^(2t) /2)(2t)dt+C_1      =(t^2 /2)e^(2t) −∫te^(2t) dt+C_1      =(t^2 /2)e^(2t) −[((te^(2t) )/2)−∫ (e^(2t) /2)dt]+C_1    I=(t^2 /2)e^(2t) −((te^(2t) )/2)+(e^(2t) /4)+C_1 +C_2    I=(e^(2t) /4)(2t^2 −2t+1)+C   I=(x^2 /4)[2(ln x)^2 −2(ln x)+1]+C .

$${let}\:\mathrm{ln}\:{x}={t}\:\:\:\Rightarrow\:{x}={e}^{{t}} \\ $$$${and}\:\:\:{dx}={e}^{{t}} {dt} \\ $$$$\:{I}=\int\left[{x}\left(\mathrm{ln}\:{x}\right)^{\mathrm{2}} \right]{dx} \\ $$$$\:\:\:=\:\int{e}^{{t}} \left({t}^{\mathrm{2}} \right)\left({e}^{{t}} {dt}\right)=\int{t}^{\mathrm{2}} {e}^{\mathrm{2}{t}} {dt} \\ $$$$\:\:\:=\:\frac{{t}^{\mathrm{2}} }{\mathrm{2}}{e}^{\mathrm{2}{t}} −\int\left(\frac{{e}^{\mathrm{2}{t}} }{\mathrm{2}}\left(\mathrm{2}{t}\right){dt}+{C}_{\mathrm{1}} \right. \\ $$$$\:\:\:=\frac{{t}^{\mathrm{2}} }{\mathrm{2}}{e}^{\mathrm{2}{t}} −\int{te}^{\mathrm{2}{t}} {dt}+{C}_{\mathrm{1}} \\ $$$$\:\:\:=\frac{{t}^{\mathrm{2}} }{\mathrm{2}}{e}^{\mathrm{2}{t}} −\left[\frac{{te}^{\mathrm{2}{t}} }{\mathrm{2}}−\int\:\frac{{e}^{\mathrm{2}{t}} }{\mathrm{2}}{dt}\right]+{C}_{\mathrm{1}} \\ $$$$\:{I}=\frac{{t}^{\mathrm{2}} }{\mathrm{2}}{e}^{\mathrm{2}{t}} −\frac{{te}^{\mathrm{2}{t}} }{\mathrm{2}}+\frac{{e}^{\mathrm{2}{t}} }{\mathrm{4}}+{C}_{\mathrm{1}} +{C}_{\mathrm{2}} \\ $$$$\:{I}=\frac{{e}^{\mathrm{2}{t}} }{\mathrm{4}}\left(\mathrm{2}{t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{1}\right)+\boldsymbol{{C}} \\ $$$$\:{I}=\frac{{x}^{\mathrm{2}} }{\mathrm{4}}\left[\mathrm{2}\left(\mathrm{ln}\:{x}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{ln}\:{x}\right)+\mathrm{1}\right]+\boldsymbol{{C}}\:. \\ $$

Commented by tawa tawa last updated on 30/May/17

I really apprecate sir. God bless you sir.

$$\mathrm{I}\:\mathrm{really}\:\mathrm{apprecate}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com