For a smal change in x, dx:

ds² = dx² + dy²

ds = sqrt [(dx² + dy²)]

s = INTEGRAL of sqrt [(dx² + dy²)]

s = INTEGRAL of sqrt [(dx² + dy² * dx²/dx²)]

s = INTEGRAL of sqrt[(1 + dy² * 1/dx²)] dx

s = INTEGRAL of sqrt[(1 + (dy/dx)²)] dx

Yes; that is one way.

in the third step, how did you integrate the right side with no delta-variable?

There is a delta variable dx. You must compute and insert dy/dx into the integrand to get the resulting arc length

To obtain the expression s = ∫√(1 + (dy/dx)²) dx, you started with the differential equation ds² = dx² + dy² and used algebraic manipulations to isolate ds. Then, you took the square root on both sides to get ds = √(dx² + dy²).

Next, you applied the integral sign (∫) to both sides of the equation ds = √(dx² + dy²), transforming it into the integral expression s = ∫√(dx² + dy²).

To simplify the expression further, you substituted dx² with (dx²/dx²), which gives s = ∫√(dx² + dy² * dx²/dx²).

Then, you simplified the expression by canceling out the dx² in the denominator of dy² * (dx²/dx²), resulting in s = ∫√(dx² + dy² * 1), which simplifies to s = ∫√(1 + (dy/dx)²) dx.

This final expression, s = ∫√(1 + (dy/dx)²) dx, represents the integral of the square root of the sum of 1 and the square of the derivative of y with respect to x, with respect to x.