Show that $\int_{0}^{4e}x^2{\ln{x}\ln{(4e-x)}\over \sqrt{x(4e-x)}}dx=-e^2\pi(\pi-1)(\pi+1)$ 
Show that
  $$\int_{0}^{4e}x^2{\ln{x}\ln{(4e-x)}\over \sqrt{x(4e-x)}}\mathrm dx=-e^2\pi(\pi-1)(\pi+1)\tag1$$

My work. Let $x=4e\sin^2{t}$ then $\mathrm dx=8e\sin{t}\cos{t}=4e\sin{(2t)}\mathrm dt$. After a long simplification we got 
$$32e^2\int\sin^4{t}\ln(4e\sin^2{t})\ln(4e\cos^2{t})\mathrm dt\tag2$$
$(2)$ probably a long IBP.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[10px,#ffd]{\ds{%
\int_{0}^{4\expo{}}x^{2}\,{\ln\pars{x}\ln\pars{4\expo{} - x} \over
\root{x\pars{4\expo{} - x}}}\,\dd x}}
\,\,\,\stackrel{x/\pars{4\expo{}}\ \mapsto\ x}{=}\,\,\,
\int_{0}^{1}\pars{4\expo{}}^{2}x^{2}\,
{\ln\pars{4\expo{}x}\ln\pars{4\expo{} - 4\expo{}x} \over
\root{4\expo{}x\pars{4\expo{} - 4\expo{}x}}}\,4\expo{}\,\dd x
\\[5mm] = &\
16\expo{}^{2}\int_{0}^{1}x^{2}\,{\bracks{\ln\pars{4\expo{}} + \ln\pars{x}}
\bracks{\ln\pars{4\expo{}} + \ln\pars{1 - x}} \over \root{x\pars{1 - x}}}\,\dd x
\\[5mm] = &\
16\expo{}^{2}\left[%
\ln^{2}\pars{4e}\!\!\int_{0}^{1}\!\!x^{3/2}\pars{1 - x}^{-1/2}\,\dd x +
\ln\pars{4e}\!\!\int_{0}^{1}\!\!{x^{2}\ln\pars{x} \over \root{x\pars{1 - x}}}
\,\dd x +
\ln\pars{4e}\!\!\int_{0}^{1}\!\!{x^{2}\ln\pars{1 - x} \over
\root{x\pars{1 - x}}}\,\dd x\right. 
\\[2mm] & \phantom{AAAAA}+
\left.\int_{0}^{1}\!\!{x^{2}\ln\pars{x}\ln\pars{1 - x} \over
\root{x\pars{1 - x}}}\,\dd x\right]
\end{align}

Note that
  $\ds{\mc{I}\pars{\mu,\nu} \equiv \int_{0}^{1}x^{\mu}\pars{1 - x}^{\nu}
\,\dd x\quad}$ 
  is equal to
  $\ds{\quad{\Gamma\pars{\mu + 1}\Gamma\pars{\mu + 1} \over
\Gamma\pars{\mu + \nu + 2}}}$ such that

\begin{align}
&\bbox[10px,#ffd]{\ds{%
\int_{0}^{4\expo{}}x^{2}\,{\ln\pars{x}\ln\pars{4\expo{} - x} \over
\root{x\pars{4\expo{} - x}}}\,\dd x}}
\\[5mm] = &\
16\expo{}^{2}\left[%
\ln^{2}\pars{4\expo{}}\,\mc{I}\pars{{3 \over 2},-\,{1 \over 2}} +
\ln\pars{4\expo{}}\,\left.\partiald{\mc{I}\pars{\mu,-1/2}}{\mu}
\right\vert_{\ \mu\ =\ 3/2} +
\ln\pars{4\expo{}}\,\left.\partiald{\mc{I}\pars{3/2,\nu}}{\nu}
\right\vert_{\ \nu\ =\ -1/2}\right.
\\[2mm] & \phantom{AAAAA}+
\left.{\left.{\partial^{2}\mc{I}\pars{\mu,\nu} \over \partial\mu\,\partial\nu}\right\vert_{\ \mu\ =\ 3/2\,,\ \nu\ =\ -1/2}}\right]
\label{1}\tag{1}
\end{align}
$$
\vphantom{\Huge A^{A^{A}}}\mbox{}
$$
\begin{equation}
\mbox{with}\quad\left\{\begin{array}{rcl}
\ds{\mc{I}\pars{{3 \over 2},-\,{1 \over 2}}} & \ds{=} &
\ds{\phantom{-}{3\pi \over 8}}
\\[5mm]
\ds{\left.\partiald{\mc{I}\pars{\mu,-1/2}}{\mu}
\right\vert_{\ \mu\ =\ 3/2}} & \ds{=} &
\ds{\phantom{-}{7\pi \over 16} - {3\pi\ln\pars{2} \over 4}}
\\[5mm]
\ds{\left.\partiald{\mc{I}\pars{3/2,\nu}}{\nu}
\right\vert_{\ \nu\ =\ -1/2}} & \ds{=} &
\ds{-\,{9\pi \over 16} - {3\pi\ln\pars{2} \over 4}}
\\[5mm]
\ds{\left.{\partial^{2}\mc{I}\pars{\mu,\nu} \over \partial\mu\,\partial\nu}\right\vert_{\ \mu\ =\ \nu\ =\ -1/2}} & \ds{=} &
\ds{-\,{3\pi \over 16} - {\pi^{3} \over 16} + {\pi\ln\pars{2} \over 4} +
{3\pi\ln^{2}\pars{2} \over 2}}
\end{array}\right.
\label{2}\tag{2}
\end{equation}

\eqref{1} and \eqref{2} lead to

$$
\bbx{\int_{0}^{4\expo{}}x^{2}\,{\ln\pars{x}\ln\pars{4\expo{} - x} \over
\root{x\pars{4\expo{} - x}}}\,\dd x =
-\expo{}^{2}\pi\pars{\pi - 1}\pars{\pi + 1}}
$$
A: Hint. Let $x=2e(1-\cos(t))$, then the given integral becomes
$$I=4e^2\int_{0}^{\pi}(1-\cos(t))^2\ln(2e(1-\cos(t)))\ln(2e(1+\cos(t)))dt.$$
Now by using the result of your previous question in Show that $\int_{0}^{2e}{\ln{(x^2)}\ln{(4e-x)}\over \sqrt{x(4e-x)}} dx=\pi(1-\zeta(2))$ , and symmetry, we get
$$\begin{align}
I&=4e^2\int_{0}^{\pi}\left(\frac{3}{2}-2\cos(t)+\frac{\cos(2t)}{2}\right)\ln(2e(1-\cos(t)))\ln(2e(1+\cos(t)))dt\\
&=e^2\pi(6−\pi^2)+0+2e^2\int_{0}^{\pi}\cos(2t)\ln(2e(1-\cos(t)))\ln(2e(1+\cos(t)))dt.\end{align}$$
Can you take it from here? For the last integral use the same Fourier series given in Show that $\int_{0}^{2e}{\ln{(x^2)}\ln{(4e-x)}\over \sqrt{x(4e-x)}} dx=\pi(1-\zeta(2))$ and the fact that $$2\cos(2t)\cos(kt)=\cos((k-2)t)+\cos((k+2)t).$$
