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Here is my plan:Meden Agan said:Hope you'll be able to outline how to go on. For now I have no idea how to do it.
\begin{align*}
I&=2\int_{\sqrt{9-2\sqrt{8}}}^3 \left(p(y)\log\left|\sqrt{1+q(y)^2}+q(y)\right|\right) \,dy\\
&=2\int_{y=\sqrt{9-2\sqrt{8}}}^{y=3} \underbrace{\left[\dfrac{p(y)}{q'(y)}\right]}_{=u}\underbrace{\log\left|\sqrt{1+z^2}+z\right|}_{=v'}\,dz\\
&=2\left[\dfrac{p(y)}{q'(y)}\cdot\left(z\log\left(\sqrt{z^2+1}+z\right)-\sqrt{z^2+1}\right)\right]_{y=\sqrt{9-2\sqrt{8}}}^{y=3}\qquad\qquad \longrightarrow C\\
&- 2\int_{y=\sqrt{9-2\sqrt{8}}}^{y=3}\left( \left(z\log\left(\sqrt{z^2+1}+z\right)-\sqrt{z^2+1}\right)\dfrac{d}{dz}\dfrac{p(y)}{q'(y)}\right)\,dz\\
&=C+2\int_{y=\sqrt{9-2\sqrt{8}}}^{y=3}\left(\sqrt{z^2+1}\dfrac{d}{dz}\dfrac{p(y)}{q'(y)}\,dz\right)-2\int_{y=\sqrt{9-2\sqrt{8}}}^{y=3}z\log\left(\sqrt{z^2+1}+z\right)\,dz
\end{align*}
The second integral has a closed form; ##C## can be calculated, so we are left with the first integral. I carried the variable within the boundaries to avoid the mistake of not adjusting them. ##q'=dq(y)/dy=dz/dy.##
But I admit that this might again result in an inconvenient polynomial.
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