\( \int _{}^{}y d x=x^{2}lnx \)

To tackle the integration \( \int y \, dx = x^2 \ln x \), it's important to remember that differentiation can be our best friend! If we differentiate \( x^2 \ln x \), we can verify the original function \( y \). Applying the product rule will get us there: Using \( u = x^2 \) and \( v = \ln x \), we differentiate to find \( \frac{d}{dx}(x^2 \ln x) = x^2 \cdot \frac{1}{x} + \ln x \cdot 2x = x + 2x \ln x \). Voila, the function \( y \) can be expressed as \( y = x + 2x \ln x \). Integrating isn’t just a math magic trick; it’s foundational in physics! Whether it’s finding the area under a curve or calculating work done by a variable force, the concepts extend the fun of integration into the real world. By linking integrals to tangible experiences, you're not just crunching numbers, you're decoding the universe!