Autonomous Equation by Separation

$$f'(x) = \frac{1}{1+[f(x)]^2} \implies (1+f^2),df = dx$$

Integrate both sides:

$$\int(1+f^2),df = \int dx \implies f + \frac{f^3}{3} = x + C$$

Using $$f(0) = 0$$: $$C = 0$$. So:

$$x = f(x) + \frac{[f(x)]^3}{3} \quad \cdots (1)$$

$$f^{-1}(t) = t + \frac{t^3}{3} \quad \cdots (2)$$

Identify $$a$$ and $$f(a)$$ for the geometric identity.

The key identity for any bijective $$f$$ with $$f(0) = 0$$ is:

$$\int_0^a f(x)dx + \int_0^{f(a)} f^{-1}(x)dx = a \cdot f(a)$$

This follows from the geometric interpretation — the two integrals together cover the rectangle of sides $$a$$ and $$f(a)$$.

$$a = \frac{4}{3} \quad \text{and} \quad f(a) = 1$$

$$\int_0^{4/3} f(x)dx + \int_0^{1} f^{-1}(x)dx = a \cdot f(a) = \frac{4}{3} \times 1 = \frac{4}{3}$$

$$\boxed{(B) \quad \frac{4}{3}}$$

Using substitution $$u = x^2 + y^2$$:
$$\frac{du}{dx} = 2x + 2y\frac{dy}{dx}$$

$$y\frac{dy}{dx} = \frac{1}{2}\frac{du}{dx} - x$$

Substituting into the original equation:
$$\frac{1}{2}\frac{du}{dx} - x + x = \sqrt{u}$$

$$\frac{1}{2}\frac{du}{dx} = \sqrt{u}$$

$$\frac{du}{dx} = 2\sqrt{u}$$

Separating variables:
$$\frac{du}{\sqrt{u}} = 2dx$$

$$2\sqrt{u} = 2x + C$$

$$\sqrt{u} = x + \frac{C}{2}$$

Applying the initial condition at $$x = 0, y = 1$$: $$u = 1$$
$$\sqrt{1} = 0 + \frac{C}{2}$$
$$C = 2$$

Therefore:
$$\sqrt{x^2 + y^2} = x + 1$$

$$x^2 + y^2 = (x+1)^2 = x^2 + 2x + 1$$

$$y^2 = 2x + 1$$