Let's tackle the integrals one by one.

1. $\int (x - 2)^2 \, dx$

First, expand the square: $(x - 2)^2 = x^2 - 4x + 4$ Now integrate term by term: $\int x^2 \, dx = \frac{x^3}{3}, \quad \int -4x \, dx = -2x^2, \quad \int 4 \, dx = 4x$ Thus, the integral becomes: $\frac{x^3}{3} - 2x^2 + 4x + C$

2. $\int \frac{2x^2 + x + 1}{x^3} \, dx$

Rewrite the integrand by splitting it into separate terms: $\frac{2x^2}{x^3} + \frac{x}{x^3} + \frac{1}{x^3} = \frac{2}{x} + \frac{1}{x^2} + \frac{1}{x^3}$ Now integrate term by term: $\int \frac{2}{x} \, dx = 2\ln|x|, \quad \int \frac{1}{x^2} \, dx = -\frac{1}{x}, \quad \int \frac{1}{x^3} \, dx = -\frac{1}{2x^2}$ Thus, the integral becomes: $2\ln|x| - \frac{1}{x} - \frac{1}{2x^2} + C$

3. $\int \frac{1 + \sqrt{x}}{x} \, dx$

First, split the integral: $\int \frac{1}{x} \, dx + \int \frac{\sqrt{x}}{x} \, dx$ The first term is straightforward: $\int \frac{1}{x} \, dx = \ln|x|$ For the second term, rewrite $\frac{\sqrt{x}}{x} = x^{-\frac{1}{2}}$, and integrate: $\int x^{-\frac{1}{2}} \, dx = 2x^{\frac{1}{2}} = 2\sqrt{x}$ Thus, the integral becomes: $\ln|x| + 2\sqrt{x} + C$

4. $\int 3(x - 2)^9 \, dx$

Use substitution. Let $u = x - 2$, so $du = dx$, and the integral becomes: $3\int u^9 \, du = 3 \cdot \frac{u^{10}}{10} = \frac{3}{10}(x - 2)^{10}$ Thus, the integral is: $\frac{3}{10}(x - 2)^{10} + C$

5. $\int x(5x^2 + 2)^9 \, dx$

Use substitution again. Let $u = 5x^2 + 2$, so $du = 10x \, dx$, and the integral becomes: $\frac{1}{10} \int u^9 \, du = \frac{1}{10} \cdot \frac{u^{10}}{10} = \frac{1}{100}u^{10}$ Substitute back $u = 5x^2 + 2$: $\frac{1}{100}(5x^2 + 2)^{10} + C$

Would you like more details on any of these solutions or further explanation on specific steps?

Related questions:

1. How do you apply substitution in integrals?
2. What are the common methods for integrating rational functions?
3. How can integrals involving powers of binomials be simplified?
4. Can you explain the use of natural logarithms in integration?
5. What is the general process for integrating functions involving square roots?

Tip: When integrating complex functions, substitution is often helpful to simplify the expression before integration.