I need the integral to calculate the arc length of the curve $8x=y^4+\frac{2}{y^2}$ from $y=1$ to $y=2$.
For now I only need to set up the integral with the formula
$$ L=\int _a^b\sqrt{1+(y')^2}dx$$
but the function that I have is equal to $x$ like this: $$x=\frac{y^4+\frac{2}{y^2}}{8}$$ As a next step I have made the derivative of that function with respect to $y$ and the result is the following: $$(f')=\frac{y^6-1}{2y^3}$$
I need help on how to create the integral with its limits of integration having the function already derived as $dy$
The reason why I only need to define the integral is because I am going to solve it with a numerical method in matlab
Note the arclength formula can be written as the following forms,
$$ L=\int_A^B\sqrt{(dx)^2+(dy)^2}=\int_{x_A}^{x_B}\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx=\int_{y_A}^{y_B} \sqrt{1+\left(\frac{dx}{dy}\right)^2}dy$$
In this problem you need to use the last expression, and plug in your derivative
$$\frac{dx}{dy}=\frac{y^6-1}{2y^3}$$