Find the area bounded by the curve y = x/((1 + x²)³)

Find the area bounded by the curve $y = \frac{x}{(1 + x^2)^3}$ and its asymptote.

We use odd symmetry and a substitution to reduce this to a convergent improper integral.

Step 1: Find the asymptote

As $x\to\pm\infty$, the denominator's sixth-power growth dominates, so $y=\frac{x}{(1+x^2)^3}\to0$. The asymptote is $y=0$.

Step 2: Analyze symmetry

\[y(-x) = \frac{-x}{(1+x^2)^3} = -y(x)\]

The function is odd, so the curve is symmetric about the origin; $y>0$ for $x>0$ and $y<0$ for $x<0$.

Step 3: Set up the area integral

\[A = \int_{-\infty}^{\infty}|y|\,dx = 2\int_0^{\infty}\frac{x}{(1+x^2)^3}\,dx\]

Step 4: Evaluate via substitution

Let $u=1+x^2$, $du=2x\,dx$. When $x=0$: $u=1$; when $x\to\infty$: $u\to\infty$.

\[\int_0^{\infty}\frac{x}{(1+x^2)^3}\,dx = \frac12\int_1^{\infty}u^{-3}\,du = \frac12\left[-\frac{1}{2u^2}\right]_1^{\infty} = \frac12\left(0+\frac12\right)=\frac14\]

Step 5: Calculate the total area

\[A = 2\cdot\frac14 = \frac12\]

\[\boxed{A = \frac12 \text{ square unit}}\]

Step 6: Verification via the antiderivative

\[\int\frac{x}{(1+x^2)^3}\,dx = -\frac{1}{4(1+x^2)^2}+C\]

\[\int_0^{\infty}\frac{x}{(1+x^2)^3}\,dx = \left[-\frac{1}{4(1+x^2)^2}\right]_0^{\infty} = 0-\left(-\frac14\right)=\frac14 \quad\checkmark\]

Summary

\[\boxed{A = \frac12 \text{ square unit}}\]

Key insight: Despite the curve extending over the whole real line, the sixth-power decay in the denominator is fast enough that the total area pinched between the curve and its asymptote converges to the simple value $\frac12$.