Find the area bounded by the curve y = 1/((1 + x²)²)

Find the area bounded by the curve $y = \frac{1}{(1 + x^2)^2}$, the line $x = 1$, and the positive part of the $x$-axis.

We will use the substitution $x=\tan\theta$ and verify against a known antiderivative.

Step 1: Set up the integral

\[A = \int_0^1 \frac{1}{(1 + x^2)^2} \, dx\]

Step 2: Use trigonometric substitution

Let $x = \tan\theta$, so $dx = \sec^2\theta \, d\theta$, and $1+x^2=\sec^2\theta$. When $x=0$: $\theta=0$. When $x=1$: $\theta=\frac\pi4$.

\[A = \int_0^{\pi/4} \frac{\sec^2\theta}{\sec^4\theta} \, d\theta = \int_0^{\pi/4} \cos^2\theta \, d\theta\]

Step 3: Use the half-angle formula

\[A = \int_0^{\pi/4} \frac{1 + \cos 2\theta}{2} \, d\theta = \frac{1}{2}\left[\theta + \frac{\sin 2\theta}{2}\right]_0^{\pi/4}\]

Step 4: Evaluate at the limits

At $\theta=\frac\pi4$: $\frac{\pi}{4} + \frac{\sin(\pi/2)}{2} = \frac\pi4+\frac12$. At $\theta=0$: $0$.

\[A = \frac{1}{2}\left(\frac{\pi}{4} + \frac{1}{2}\right) = \frac{\pi}{8} + \frac{1}{4}\]

\[\boxed{A = \frac{\pi}{8} + \frac{1}{4} = \frac{\pi + 2}{8} \text{ square units}}\]

Step 5: Numerical value

\[A \approx 0.643 \text{ square units}\]

Verification via the closed-form antiderivative

\[\int \frac{1}{(1 + x^2)^2} \, dx = \frac{1}{2}\left(\arctan x + \frac{x}{1 + x^2}\right) + C\]

\[A = \left[\frac{1}{2}\left(\arctan x + \frac{x}{1 + x^2}\right)\right]_0^1 = \frac{1}{2}\left(\frac{\pi}{4}+\frac12\right) - 0 = \frac{\pi}{8}+\frac14 \quad \checkmark\]

Summary

\[\boxed{A = \frac{\pi + 2}{8} \approx 0.643 \text{ square units}}\]

Key insight: $\frac{1}{(1+x^2)^2}$ is closely tied to the derivative of $\arctan x$, and the area combines a transcendental term ($\pi$) with a rational one (2) in the compact form $\frac{\pi+2}{8}$.