Vtyjkyuk

The following answer to the question $$int z(ln z)^2, dz $$ is given by

$$frac12(z^2(ln z)^2-z^2(ln z)+ fracz^22)+ C. $$

I first used a substitution of $u=z ln z$ as $ln z$ cannot be integrated cleanly (and so Integration by parts is inapplicable), reducing the integral to $$int z ln z (ln z+1) , dz - int z ln z , dz.$$

Still, $ln z$ cannot be integrated cleanly so did a 2nd substitution of $y= ln z$ which will reduce the integral(after applying both substitutions) into $$int u,du - int y e^2y , dy$$ Then I proceeded with Integration by parts.

If the power of $ln z$ was any larger, it would be tedious to use n substitutions and then proceed with integration by parts. Does anyone have an alternative solution to this problem?

$$I=displaystyleintxln^2left(xright),mathrmdx$$
Integration by parts

$$I=dfracx^2ln^2left(xright)2-displaystyleintxlnleft(xright),mathrmdx$$

$$I_1=displaystyleintxlnleft(xright),mathrmdx$$
again apply integration by parts

$$I_1=dfracx^2lnleft(xright)2-displaystyleintdfracx2,mathrmdx=dfracx^2lnleft(xright)2-dfracx^24$$

$$I=dfracx^2ln^2left(xright)2-dfracx^2lnleft(xright)2+dfracx^24+C$$

With substitution $ln z=u$ or $z=e^u$ then
$$I=int zln^2z dz=int e^2uu^2 du$$
let $2u=w$ and use the formula $displaystyleint e^tp(t)dt=e^t(p-p'+p''-cdots)+C$ where $p$ is a polynomial therefore
$$I=dfrac18int e^ww^2 dw=dfrac18e^wleft(w^2-2w+2right)+C=colorbluedfrac18z^2left(4ln^2z-4ln z+2right)+C$$