Create the Euler–Lagrange equation for the following questions (if it's necessary change the variables).
$$\tag{1}\int_{y_1}^{y_2}\dfrac{x'^2}{\sqrt{x'^2+x^2}}\,\mathrm{d}y$$
$$\tag{2}\int_{x_1}^{x_2}y^{3/2}\,\mathrm{d}s$$
$$\tag{3}\int\dfrac{y\cdot{y'}}{1+yy'}\,\mathrm{d}x$$
I don't have an idea about $(1)$ and $(3)$. But here it's what I have tried for $(2)$.
2) $$\int_{x_1}^{x_2}y^{3/2}\,\mathrm{d}s = \int_{x_1}^{x_2}y^{3/2}(1+y'^2)^{1/2} = \int_{y_1}^{y_2}(1+x'^2)(y^{3/2})\,\mathrm{d}y$$
So our Euler equation is:
$$\dfrac{\mathrm{d}}{\mathrm{d}y}\left(\dfrac{\partial F}{\partial x'}\right) - \dfrac{\partial F}{\partial x} = 0$$
Then I have to find $Y'$ or $X'$. But I did not take a differential equations course yet, we use Beltrami identity to calculate the extremum points.
We know that the functional
$$I[y]=\int_{x_1}^{x_2} f(x,y(x),y'(x))\,\mathrm{d}x$$
is extremized only if $y$ satisfies the Euler–Lagrange equation
$$\frac{\partial{f}}{\partial{y}}-\frac{\mathrm{d}}{\mathrm{d}x}\frac{\partial{f}}{\partial{y'}}=0.$$
Now, if $f$ is independent of $x$, then the Euler–Lagrange equation reduces to the Beltrami identity
$$f-y'\frac{\partial{f}}{\partial{y'}}=C=\text{constant}.$$
Note that the names of the variables are immaterial. Henceforth, we use the same notation as above.
In the first case, we have $f(x,y,y')=\dfrac{y'^2}{\sqrt{y^2+y'^2}}$, which is independent of $x$. Hence we can use the Beltrami identity. Differentiation gives
$$\frac{\partial{f}}{\partial{y'}}=\frac{2y^2y'+y'^3}{\left(y^2+y'^2\right)^{3/2}}\implies\frac{y'^2}{\sqrt{y^2+y'^2}}-\frac{2y^2y'^2+y'^4}{\left(y^2+y'^2\right)^{3/2}}=C.$$
Rewriting the first term gives
$$\frac{y^2y'^2+y'^4}{\left(y^2+y'^2\right)^{3/2}}-\frac{2y^2y'^2+y'^4}{\left(y^2+y'^2\right)^{3/2}}=C\implies-\frac{y^2y'^2}{\left(y^2+y'^2\right)^{3/2}}=C,$$
which is a first-order non-linear ordinary differential equation. In total, two constants will be obtained.
Of course, we could use the Euler–Lagrange equation instead, but that would be much more complicated; just consider the derivatives:
$$\frac{\partial{f}}{\partial{y}}=-\frac{yy'^2}{\left(y^2+y'^2\right)^{3/2}},\,\frac{\partial{f}}{\partial{y'}}=\frac{2y^2y'+y'^3}{\left(y^2+y'^2\right)^{3/2}},\,\frac{\mathrm{d}}{\mathrm{d}x}\frac{\partial{f}}{\partial{y'}}=\frac{y(2y^2-y'^2)(yy''-y'^2)}{\left(y^2+y'^2\right)^{5/2}}.$$
The second and thrid cases can be treated similarly; both integrands are independent of $x$. However, in the second case, note that $\mathrm{d}s=\sqrt{1+y'^2}\,\mathrm{d}x$, so $f(x,y,y')=y^{3/2}\sqrt{1+y'^2}$; this expression is independent of $x$, as it should be.