1. **Stating the problem:** Calculate the definite integral $$\int_0^1 \left(x^2 + \frac{2}{x^2}\right) dx$$. 2. **Formula and rules:** The integral of a sum is the sum of the integrals: $$\int (f(x) + g(x)) dx = \int f(x) dx + \int g(x) dx$$ Also, for power functions: $$\int x^n dx = \frac{x^{n+1}}{n+1} + C, \quad n \neq -1$$ 3. **Break the integral:** $$\int_0^1 x^2 dx + \int_0^1 \frac{2}{x^2} dx = \int_0^1 x^2 dx + 2 \int_0^1 x^{-2} dx$$ 4. **Calculate each integral:** - For $$\int_0^1 x^2 dx$$: $$\int x^2 dx = \frac{x^{3}}{3}$$ Evaluate from 0 to 1: $$\left. \frac{x^{3}}{3} \right|_0^1 = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3}$$ - For $$2 \int_0^1 x^{-2} dx$$: $$\int x^{-2} dx = \int x^{-2} dx = \frac{x^{-1}}{-1} = -x^{-1} = -\frac{1}{x}$$ Evaluate from 0 to 1: $$2 \left( \left. -\frac{1}{x} \right|_0^1 \right) = 2 \left(-1 + \lim_{x \to 0^+} \frac{1}{x} \right)$$ 5. **Check for convergence:** The term $$\lim_{x \to 0^+} \frac{1}{x}$$ diverges to infinity, so the integral $$\int_0^1 \frac{2}{x^2} dx$$ diverges. 6. **Conclusion:** Since one part of the integral diverges, the entire integral $$\int_0^1 \left(x^2 + \frac{2}{x^2}\right) dx$$ diverges and does not have a finite value.