Til the last breath-2

$\color{#302B94}{\begin{aligned} &\int_{0}^1\left( 3r^2y^2+4ry+r^2\right)\,dy\\&=(r^2y^3+2r^2y^2+r^2y)|_{0}^{1}\\& =2(r^2+r)\end{aligned}}$

---

$\color{#302B94}{\displaystyle\sum_{r=1}^{x}\left(\int_{0}^1\left( 3r^2y^2+4ry+r^2\right)\,dy\right)\\ =\displaystyle\sum_{r=1}^{x}\left(2r(r+1)\right) \\=2(r^2+r)}$ Using formula for sum of n,n $^2$ expression can be simplified as: $\color{#302B94}{=2\left(\dfrac{x(x+1)(2x+1)}{6}+\dfrac{x(x+1)}{2}\right)~~~\\=\dfrac{2x(x+1)(x+2)}{3}}$

---

$\color{#302B94}{\mathfrak{T}=\int_{2}^{\color{forestgreen}{\phi}}\dfrac{\cancel{x(x+1)}}{\frac{2\cancel{x(x+1)}(x+2)}{3}}\, dx}\,$ $\Large\color{#302B94}{=\dfrac 32(\ln(\color{forestgreen}{\dfrac{\phi+2}{4}}))}$

---

$\Large\color{#3D99F6}{\psi}=\color{#624F41}{8^{\small{\frac 12+\frac18+\frac{1}{32}}}\cdot 216^{\small{\frac 14+\frac{1}{16}+\frac{1}{64}}}\\=8^{\small{\frac 23}}\cdot 216^{\small{\frac 13}}=24}$

$\Large \color{forestgreen}{\phi}=\dfrac{\color{#3D99F6}{\psi}}{4}+1=\color{forestgreen}{7}$

---

Hence, $\Large \color{#EC7300}{\mathfrak T}=\color{#EC7300}{\dfrac 32\ln \dfrac 94=\ln \dfrac{3^3}{8}}$

$\Huge\therefore~\color{#0C6AC7}{ \varphi=}\color{#456461}{\boxed{\color{#EC7300}{\boxed{\color{#0C6AC7}{\mathbf{3}}}}}}$