For $x>0$, one may check that
\begin{align}
\small{\sqrt {x+\frac{1}{\sqrt{x}}}}&=\sqrt{x}+\frac {1}{2x + \frac 1{2\sqrt{x} +\frac{1}{2x+\frac{1}{\sqrt{x}+\sqrt {x+\frac{1}{\sqrt{x}}}}}}} \tag1
\end{align}
then, using
\begin{align}
\small{\sqrt{x}+\sqrt {x+\frac{1}{\sqrt{x}}}>2\sqrt{x}}
\end{align}
one gets
$$\int_1^8\frac{dx}{\sqrt {x+\frac{1}{\sqrt{x}}}} <\int_1^8\frac{dx}{\sqrt{x}+\frac {1}{2x + \frac 1{2\sqrt{x} +\frac{1}{2x+\frac{1}{2\sqrt{x}}}}}}. \tag2
$$
By the change of variable $u=2\sqrt{x}$, $du=\frac{dx}{\sqrt{x}}$, setting $\varphi=\frac{1+\sqrt{5}}2$, one has
\begin{align}
&\int_1^8\frac{dx}{\sqrt{x}+\frac {1}{2x + \frac 1{2\sqrt{x} +\frac{1}{2x+\frac{1}{2\sqrt{x}}}}}}
\\\\&=\int_1^{8}\frac{16x^{3}+12x^{3/2}+1}{16x^{3}+20x^{3/2}+5}\:\frac{dx}{\sqrt{x}}
\\\\&=\int_2^{4\sqrt{2}}\frac{u^6+6u^3+4}{u^6+10u^3+20}\:du, \qquad u=2^{1/3}5^{1/6}t,
\\\\&=4\sqrt{2}-2-2^{4/3}5^{-1/3}\varphi \int_{2^{2/3}5^{-1/6}}^{2^{13/6}5^{-1/6}}\frac{dt}{t^3+\varphi}-\frac{2^{4/3}5^{2/3}}{\varphi}\int_{2^{2/3}5^{-1/6}}^{2^{13/6}5^{-1/6}}\frac{dt}{t^3+\frac1{\varphi}}.
\end{align}
Then, using the evaluation
$$
\int_0^b\frac{dt}{t^3+c^3}=\frac{\sqrt{3}\pi}{18c^2}+\frac{\sqrt{3}}{3c^2}\arctan \left(\frac{2b-c}{\sqrt{3}c}\right)-\frac{1}{6c^2}\ln \left(\frac{(2b-c)^2+3c^2}{4(b+c)^2}\right),\,c>0,\,b>0,
$$ one gets
$$
\int_1^8\frac{dx}{\sqrt{x}+\frac {1}{2x + \frac 1{2\sqrt{x} +\frac{1}{2x+\frac{1}{2\sqrt{x}}}}}}=3.3199\color{red}{58928}\cdots. \tag3
$$
In a like manner, for $x>0$, one may check that
\begin{align}
\small{\sqrt {x+\frac{1}{\sqrt{x}}}}&=\frac{1}{x^{1/4}}+\frac {1}{\frac{2}{x^{5/4}} + \frac 1{\frac{2}{x^{1/4}} +\frac{1}{\frac{2}{x^{5/4}}+\frac{1}{\frac{1}{x^{1/4}}+\sqrt {x+\frac{1}{\sqrt{x}}}}}}} \tag4
\end{align}
then, using
\begin{align}
\small{\frac{1}{x^{1/4}}+\sqrt {x+\frac{1}{\sqrt{x}}}>\frac{2}{x^{1/4}}}
\end{align}
one gets
$$\int_0^1\frac{dx}{\sqrt {x+\frac{1}{\sqrt{x}}}} <\int_0^1\frac{dx}{\frac{1}{x^{1/4}}+\frac {1}{\frac{2}{x^{5/4}} + \frac 1{\frac{2}{x^{1/4}} +\frac{1}{\frac{2}{x^{5/4}}+\frac{x^{1/4}}2}}}}. \tag5
$$
By the change of variable $u=\sqrt[4]{x}$, $dx=4u^3du$, setting $\varphi=\frac{1+\sqrt{5}}2$, one has
\begin{align}
&\int_0^1\frac{dx}{\frac{1}{x^{1/4}}+\frac {1}{\frac{2}{x^{5/4}} + \frac 1{\frac{2}{x^{1/4}} +\frac{1}{\frac{2}{x^{5/4}}+\frac{x^{1/4}}2}}}}
\\\\&=\int_0^{1}\frac{x^{3}+12x^{3/2}+16}{5x^{3}+20x^{3/2}+16}\:\sqrt[4]{x}\:dx
\\\\&=4\int_0^1\frac{u^{12}+12u^6+16}{5u^{12}+20u^6+16}\:u^4du, \qquad u=2^{1/3}5^{-1/12}t,
\\\\&=\frac4{25}+\frac{6}{\sqrt{5}}+2^{-2/3}5^{-1/3}\varphi \int_0^{2^{-1/3}5^{1/12}}\!\!\frac{t^4dt}{t^6+\varphi}-\frac{2^{4/3}5^{-1/3}}{\varphi}\int_0^{2^{-1/3}5^{1/12}}\!\!\frac{t^4dt}{t^6+\frac1{\varphi}}.
\end{align}
Then, using the evaluation
\begin{align}
&6c\int_0^b\frac{t^4dt}{t^6+c^6}\qquad (c>0,\,b>0)
\\\\&=\arctan \left(\sqrt{3}+\frac{2b}{c}\right)-\arctan \left(\sqrt{3}-\frac{2b}{c}\right)+2\arctan \left(\frac{b}{c}\right)-\frac{\sqrt{3}}{2}\ln \left(\frac{b^2+\sqrt{3}bc+c^2}{b^2+\sqrt{3}bc+c^2}\right)
\end{align} one obtains
$$
\int_0^1\frac{dx}{\frac{1}{x^{1/4}}+\frac {1}{\frac{2}{x^{5/4}} + \frac 1{\frac{2}{x^{1/4}} +\frac{1}{\frac{2}{x^{5/4}}+\frac{x^{1/4}}2}}}}=0.6739\color{red}{21415}\cdots. \tag6
$$
Finally, combining $(2)$, $(3)$, $(5)$ and $(6)$ yields
$$\int_0^8\frac{dx}{\sqrt {x+\frac{1}{\sqrt{x}}}} <\color{red}{3.99}38803\cdots<\color{red}{4}-\frac1{2019}=\color{red}{3.99}950\cdots \tag7
$$
as desired.