TITLE: Find the integral $\int _{1}^{e} (x \ln x)^2 dx$.
QUESTION [5 upvotes]: Find : $$\int _{1}^{e} (x \ln x)^2 \;dx.$$

My answer:
I have tried integration by parts with $u = x^2$ and $dv = (\ln x)^2$ but I end up having the same integration another time!
I reversed the role of $u$ & $v$, but it also did not work?
Do you have any suggestions ?

REPLY [2 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\int_{1}^{\expo{}}x^{\nu}\,\dd x & =
\left.{x^{\nu +1} \over \nu + 1}\,\right\vert_{\ 1}^{\ \expo{}} =
{\expo{\nu +1} - 1 \over \nu + 1}
\\[5mm] &
\textsf{Derive twice respect of}\ \nu:
\\
\int_{1}^{\expo{}}x^{\nu}\ln^{2}\pars{x}\,\dd x & =
\totald[2]{}{\nu}\pars{\expo{\nu +1} - 1 \over \nu + 1} =
{\pars{\nu^{2} + 1}\expo{\nu + 1} - 2 \over \pars{\nu +1}^{3}}
\\[5mm] &
\textsf{Evaluate the limit}\ \nu \to 2:
\\
\int_{1}^{\expo{}}\bracks{x\ln\pars{x}}^{2}\,\dd x & =
\bbx{5\expo{3} - 2 \over 27} \approx 3.6455
\end{align}