The given integral is:

$\int_0^\infty \frac{x \, dx}{(1 + x^2)^2}.$

To evaluate this integral, we can proceed as follows:

Step 1: Substitution

We use the substitution: $x = \tan\theta \quad \text{and} \quad dx = \sec^2\theta \, d\theta.$

Under this substitution:

1. $x^2 = \tan^2\theta$,
2. $1 + x^2 = \sec^2\theta$,
3. $dx = \sec^2\theta \, d\theta$.

The integral limits transform as:

• When $x = 0$, $\theta = 0$,
• When $x \to \infty$, $\theta \to \frac{\pi}{2}$.

Thus, the integral becomes:

$\int_0^\infty \frac{x \, dx}{(1 + x^2)^2} = \int_0^{\pi/2} \frac{\tan\theta \cdot \sec^2\theta \, d\theta}{\sec^4\theta}.$

Step 2: Simplification

Simplify the integrand step by step:

• $\sec^4\theta$ in the denominator and $\sec^2\theta$ in the numerator cancel partially: $\frac{\tan\theta \cdot \sec^2\theta}{\sec^4\theta} = \frac{\tan\theta}{\sec^2\theta}.$

Now, since $\sec^2\theta = 1 + \tan^2\theta$, we get: $\frac{\tan\theta}{\sec^2\theta} = \frac{\tan\theta}{1 + \tan^2\theta}.$

Step 3: Substitution for Simplicity

Let $u = \tan\theta$, so:

• $du = \sec^2\theta \, d\theta$,
• As $\theta$ goes from $0$ to $\frac{\pi}{2}$, $u$ goes from $0$ to $\infty$.

Thus, the integral becomes: $\int_0^{\pi/2} \frac{\tan\theta}{1 + \tan^2\theta} \, d\theta = \int_0^\infty \frac{u}{1 + u^2} \cdot \frac{du}{1}.$

Step 4: Solving the New Integral

Simplify the integrand: $\frac{u}{1 + u^2}.$

This is a straightforward integral. We use the substitution: $v = 1 + u^2 \implies dv = 2u \, du.$

Rewriting the integral: $\int \frac{u}{1 + u^2} \, du = \frac{1}{2} \int \frac{dv}{v}.$

The integral of $\frac{1}{v}$ is $\ln v$. Therefore: $\int \frac{u}{1 + u^2} \, du = \frac{1}{2} \ln(1 + u^2) + C.$

Step 5: Applying Limits

We evaluate the integral from $0$ to $\infty$:

• At $u = 0$: $\frac{1}{2} \ln(1 + 0^2) = \frac{1}{2} \ln(1) = 0.$
• At $u \to \infty$: $\frac{1}{2} \ln(1 + u^2) \to \frac{1}{2} \ln(\infty) \to \infty.$

Thus, we see that the integral evaluates to divergence.