Question:

Step 1: Find the intersection points between the curve and the line. We need to determine where $f(x) = 1$. Setting up the equation: $$\frac{x^2-1}{x^2+1} = 1$$ Multiplying both sides by $(x^2+1)$: $$x^2 - 1 = x^2 + 1$$ This simplifies to $-1 = 1$, which is impossible. Therefore, the curve and the line never intersect. Step 2: Rewrite the function to understand its behavior relative to the line $y = 1$. We can rewrite $f(x)$ by performing algebraic manipulation: $$f(x) = \frac{x^2-1}{x^2+1} = \frac{x^2+1-2}{x^2+1} = 1 - \frac{2}{x^2+1}$$ This form reveals that $f(x) < 1$ for all real $x$. As $x \to \pm\infty$, the term $\frac{2}{x^2+1} \to 0$, so $f(x) \to 1^-$. The curve approaches the line $y = 1$ asymptotically from below. Step 3: Set up the integral for the area between the curves. The area bounded between the line $y = 1$ and the curve $f(x)$ is given by: $$A = \int_{-\infty}^{\infty} \left[1 - f(x)\right]dx$$ Substituting our expression for $f(x)$: $$A = \int_{-\infty}^{\infty} \left[1 - \left(1 - \frac{2}{x^2+1}\right)\right]dx = \int_{-\infty}^{\infty} \frac{2}{x^2+1}dx$$ Step 4: Evaluate the improper integral. We evaluate this integral using the standard antiderivative: $$A = 2\int_{-\infty}^{\infty} \frac{1}{x^2+1}dx = 2[\arctan x]_{-\infty}^{\infty}$$ Computing the limits: $$A = 2\left[\lim_{x \to \infty} \arctan x - \lim_{x \to -\infty} \arctan x\right]$$ $$A = 2\left[\frac{\pi}{2} - \left(-\frac{\pi}{2}\right)\right] = 2 \cdot \pi = 2\pi$$ **Final Answer:** The area bounded by the curve $f(x) = \dfrac{x^2-1}{x^2+1}$ and the line $y = 1$ is $\boxed{2\pi}$, which corresponds to **Option 2**.