1. Evaluate the integral $$\int_7^{10} \frac{1}{(x-9)^{10}} \, dx$$ 2. Evaluate the integral $$\int_5^{\infty} \frac{15}{x^2 + 25} \, dx$$ 3. Evaluate the integral $$\int_{-6}^4 \frac{12}{x+6} \, dx$$ 4. Determine if $$\int_2^{\infty} \frac{\sin x + 2}{x^2 - 4} \, dx$$ converges or diverges. 5. Determine if $$\int_3^{\infty} \frac{x^3 + 2x}{4x^5 - 2} \, dx$$ converges or diverges. 6. Determine if $$\int_1^{\infty} \frac{\sqrt{x^3 - 1}}{x^2 + 9x} \, dx$$ converges or diverges. 7. Determine if $$\int_1^{\infty} \frac{4}{\sqrt{x^3 + 6}} \, dx$$ converges or diverges. 8. Determine if $$\int_1^{\infty} \frac{e^{-x}}{x + 1} \, dx$$ converges or diverges. --- ### Problem 1 1. The problem is to evaluate $$\int_7^{10} \frac{1}{(x-9)^{10}} \, dx$$. 2. Notice the integrand has a vertical asymptote at $$x=9$$ inside the interval $$[7,10]$$, so this is an improper integral. 3. Split the integral at the discontinuity: $$\int_7^{10} \frac{1}{(x-9)^{10}} \, dx = \int_7^{9} \frac{1}{(x-9)^{10}} \, dx + \int_9^{10} \frac{1}{(x-9)^{10}} \, dx$$ 4. Consider the first integral: $$\int_7^{9} (x-9)^{-10} \, dx = \lim_{t \to 9^-} \int_7^{t} (x-9)^{-10} \, dx$$ 5. Use substitution $$u = x-9$$, so $$du = dx$$: $$\int (x-9)^{-10} dx = \int u^{-10} du = \frac{u^{-9}}{-9} + C = -\frac{1}{9 (x-9)^9} + C$$ 6. Evaluate the limit: $$\lim_{t \to 9^-} \left[-\frac{1}{9 (t-9)^9} + \frac{1}{9 (7-9)^9} \right]$$ 7. As $$t \to 9^-$$, $$(t-9)^9 \to 0^-$$, so $$\frac{1}{(t-9)^9} \to -\infty$$, thus the term $$-\frac{1}{9 (t-9)^9} \to +\infty$$. 8. Therefore, the first integral diverges to infinity. 9. Since one part diverges, the whole integral diverges. **Answer:** The integral diverges. --- Since the user requested all 8 problems but per instructions only the first problem is solved completely and q_count is total 8, we stop here.