Find the integral $\int _{1}^{e} (x \ln x)^2 dx$. 
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 ? 
 A: Hint: make thus substitution $y =\ln\, x$ and then integrate by parts. 
A: Hint:
Let $\ln x=y\implies x=e^y,dx=e^y\ dy$
$$\int_1^e(x\ln x)^2=\int_0^1e^{3y}y^2\ dy$$
Now $\dfrac{d(e^{my}y^n)}{dy}=me^{my}y^n+e^{my}ny^{n-1}$
If $\displaystyle I(n)=\int e^{my}y^n dy,$
$$mI(n)+nI(n-1)=e^{my}y^n+K$$
Here $m=3,n=2$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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}}}
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\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}
A: Let $u=(\ln x)^2$ and $dv=x^2 \,dx$. Then $du=2(\ln x)\frac{1}{x} \, dx$ and $v=\frac{x^3}{3}$. So
$$I=\int_1^e(x \ln x)^2 \, dx=(\ln x)^2 \frac{x^3}{3}\Big|_{1}^{e}-\frac{2}{3}\int_1^e x^2 \ln x \, dx.$$
Now we will solve the integral on the right side. Call the integral as $J$. For this $u=\ln x$ and $dv=x^2 \, dx$. So $du= \frac{1}{x}\,dx$ and $v=\frac{x^3}{3}$.Then
$$J=(\ln x) \frac{x^3}{3}\Big|_{1}^{e}-\frac{1}{3}\int_1^e x^ 2\, dx=(\ln x) \frac{x^3}{3}\Big|_{1}^{e}-\frac{x^3}{9}\Big|_{1}^{e}=\frac{e^3}{3}-\left(\frac{e^3-1}{9}\right)=\frac{2e^3+1}{9}.$$
So
$$I=\frac{e^3}{3}-\frac{2}{3}\left(\frac{2e^3+1}{9}\right)=\frac{5e^3-2}{27}$$
