Change the independent variable to simplify the Euler equation, and then find a first integral of it.

$\mathbf{1}\mathbf{.}\mathbf{}\mathbf{}{\mathbf{\int }}_{{\mathbf{x}}_{\mathbf{1}}}^{{\mathbf{x}}_{\mathbf{2}}}{\mathbf{y}}^{\raisebox{1ex}{$\mathbf{3}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{2}$}\right.}\mathbf{ds}$

The given integral is${\int }_{x_1}^{x_2}{y}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}ds$

For the integral, $\mathbf{I}\mathbf{}\left(\epsilon \right)\mathbf{=}{\mathbf{\int }}_{{\mathbf{x}}_{\mathbf{1}}}^{{\mathbf{x}}_{\mathbf{2}}}\mathbf{F}\left(x,y,y\text{'}\right)\mathbf{dx}$ the Euler equation defined asrole="math" localid="1665061125622" $\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\frac{\mathbf{d}\mathbf{F}}{\mathbf{d}\mathbf{y}\mathbf{\text{'}}}\mathbf{-}\frac{\mathbf{d}\mathbf{F}}{\mathbf{d}\mathbf{y}}\mathbf{=}\mathbf{0}$

First let’s change the integral into ideal form

$$\begin{gathered} \int y^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}ds=\int y^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\sqrt{d{x}^2+d{y}^2} \\ =\int y^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\sqrt{1+y{\text{'}}^2}dx \end{gathered}$$

Now let’s change the variables as follows

$$\begin{gathered} dx=x\text{'}dy \\ y\text{'}=\frac{1}{x\text{'}} \end{gathered}$$

By using the two equations above, we can rewrite the integral

$$\begin{gathered} \int y^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\sqrt{1+y{\text{'}}^2}dx=\int y^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\sqrt{1+y{\text{'}}^2}x\text{'}dy \\ =\int y^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\sqrt{1+x{\text{'}}^{-2}}x\text{'}dy \\ =\int y^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\sqrt{x{\text{'}}^2+}1dy \end{gathered}$$

Let $F\left(x,y,y\text{'}\right)=y^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\sqrt{1+x{\text{'}}^2}$

By the definition of Euler equation $\frac{d}{dy}\frac{\partial F}{\partial x\text{'}}-\frac{\partial F}{\partial x}=0$

Differentiate$\mathrm{F}$with respect to$\mathrm{x}\text{'}$and$\mathrm{x}$

$$\begin{gathered} \frac{\partial F}{\partial x\text{'}}=\frac{x\text{'}{y}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{\sqrt{1+x{\text{'}}^2}} \\ \frac{\partial F}{\partial x}=0 \end{gathered}$$

The Euler become

$$\begin{gathered} \frac{d}{dy}\left[\frac{x\text{'}{y}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{\sqrt{1+x{\text{'}}^2}}\right]=0 \\ \frac{x\text{'}{y}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{\sqrt{1+x{\text{'}}^2}}=C \\ x{\text{'}}^2=\frac{C^2}{y^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-C^2} \end{gathered}$$

Therefore, the differential equation is$x\text{'}=\pm \frac{C}{\sqrt{y^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-C^2}}$

Since minus can be absorbed into the constant.

$x\text{'}=\frac{C}{\sqrt{y^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-C^2}}$

Therefore, the first integral of the Euler equation is$x\text{'}=\frac{C}{\sqrt{y^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-C^2}}$