1. **State the problem:** Evaluate the integral $$\int_0^{\infty} \frac{x \ln x}{(1+x^2)^2} \, dx.$$\n\n2. **Recall the formula and approach:** This integral involves a logarithm and a rational function. A useful substitution is to let $$x = \tan \theta,$$ which transforms the integral over $x \in [0, \infty)$ to $\theta \in [0, \frac{\pi}{2})$.\n\n3. **Perform the substitution:**\n$$x = \tan \theta \implies dx = \sec^2 \theta \, d\theta,$$\nand note that $$1 + x^2 = 1 + \tan^2 \theta = \sec^2 \theta.$$\n\n4. **Rewrite the integral:**\n$$\int_0^{\infty} \frac{x \ln x}{(1+x^2)^2} \, dx = \int_0^{\frac{\pi}{2}} \frac{\tan \theta \ln(\tan \theta)}{(\sec^2 \theta)^2} \cdot \sec^2 \theta \, d\theta = \int_0^{\frac{\pi}{2}} \tan \theta \ln(\tan \theta) \cdot \frac{\sec^2 \theta}{\sec^4 \theta} \, d\theta = \int_0^{\frac{\pi}{2}} \tan \theta \ln(\tan \theta) \cdot \frac{1}{\sec^2 \theta} \, d\theta.$$\nSince $$\frac{1}{\sec^2 \theta} = \cos^2 \theta,$$ and $$\tan \theta = \frac{\sin \theta}{\cos \theta},$$ the integral becomes\n$$\int_0^{\frac{\pi}{2}} \frac{\sin \theta}{\cos \theta} \ln \left( \frac{\sin \theta}{\cos \theta} \right) \cos^2 \theta \, d\theta = \int_0^{\frac{\pi}{2}} \sin \theta \cos \theta \ln \left( \frac{\sin \theta}{\cos \theta} \right) \, d\theta.$$\n\n5. **Simplify the logarithm:**\n$$\ln \left( \frac{\sin \theta}{\cos \theta} \right) = \ln(\sin \theta) - \ln(\cos \theta).$$\nSo the integral splits into\n$$\int_0^{\frac{\pi}{2}} \sin \theta \cos \theta \ln(\sin \theta) \, d\theta - \int_0^{\frac{\pi}{2}} \sin \theta \cos \theta \ln(\cos \theta) \, d\theta.$$\n\n6. **Use symmetry:** Let\n$$I = \int_0^{\frac{\pi}{2}} \sin \theta \cos \theta \ln(\sin \theta) \, d\theta,$$\nthen by substituting $\theta = \frac{\pi}{2} - t$, we find\n$$\int_0^{\frac{\pi}{2}} \sin \theta \cos \theta \ln(\cos \theta) \, d\theta = I.$$\n\n7. **Therefore, the original integral is:**\n$$I - I = 0.$$\n\n8. **Conclusion:** The value of the integral is $$\boxed{0}.$$