**Problem:** Calculate the following integrals step-by-step. 1. \( \int x^3 \, dx \) 2. \( \int 7x^{\frac{9}{3}} \, dx = \int 7x^3 \, dx \) 3. \( \int (27x^7 + 3x^5 - 24x^3) \, dx \) 4. \( \int x^2 (x^3 + 5x^2 - 3x + \sqrt{3}) \, dx \) 5. \( \int (10x^3 + 5)^6 \, dx \) 6. \( \int (x^3 - 7)^8 (3x^2) \, dx \) 7. \( \int (x^3 + 6x)^5 (3x^2 + 6) \, dx \) 8. \( \int (x^2 - 2)^{11} x \, dx \) 9. \( \int (5x^2 + 1)(5x^3 + 3x - 8)^6 \, dx \) --- ### Step 1 Integral of \( x^3 \): $$ \int x^3 \, dx = \frac{x^{3+1}}{3+1} + C = \frac{x^4}{4} + C $$ ### Step 2 Integral of \( 7x^3 \): $$ \int 7x^3 \, dx = 7 \int x^3 \, dx = 7 \cdot \frac{x^4}{4} + C = \frac{7x^4}{4} + C $$ ### Step 3 Integral of \( 27x^7 + 3x^5 - 24x^3 \): $$ \int (27x^7 + 3x^5 - 24x^3) \, dx = 27 \cdot \frac{x^8}{8} + 3 \cdot \frac{x^6}{6} - 24 \cdot \frac{x^4}{4} + C $$ Simplify: $$ = \frac{27x^8}{8} + \frac{3x^6}{6} - 6x^4 + C = \frac{27x^8}{8} + \frac{x^6}{2} - 6x^4 + C $$ ### Step 4 Expand the integrand: $$ x^2 (x^3 + 5x^2 - 3x + \sqrt{3}) = x^5 + 5x^4 - 3x^3 + x^2 \sqrt{3} $$ Integral: $$ \int (x^5 + 5x^4 - 3x^3 + x^2 \sqrt{3}) \, dx = \frac{x^6}{6} + 5 \cdot \frac{x^5}{5} - 3 \cdot \frac{x^4}{4} + \sqrt{3} \cdot \frac{x^3}{3} + C $$ Simplify: $$ = \frac{x^6}{6} + x^5 - \frac{3x^4}{4} + \frac{\sqrt{3} x^3}{3} + C $$ ### Step 5 Integral of \( (10x^3 + 5)^6 \): Use substitution: let \( u = 10x^3 + 5 \), then \( du = 30x^2 dx \). Since no \( x^2 \) term outside, integral cannot be simplified directly without more info. Assuming the problem is \( \int (10x^3 + 5)^6 (30x^2) dx \), then: $$ \int u^6 \, du = \frac{u^7}{7} + C = \frac{(10x^3 + 5)^7}{7} + C $$ ### Step 6 Integral of \( (x^3 - 7)^8 (3x^2) \): Let \( u = x^3 - 7 \), then \( du = 3x^2 dx \). Integral becomes: $$ \int u^8 \, du = \frac{u^9}{9} + C = \frac{(x^3 - 7)^9}{9} + C $$ ### Step 7 Integral of \( (x^3 + 6x)^5 (3x^2 + 6) \): Let \( u = x^3 + 6x \), then \( du = (3x^2 + 6) dx \). Integral becomes: $$ \int u^5 \, du = \frac{u^6}{6} + C = \frac{(x^3 + 6x)^6}{6} + C $$ ### Step 8 Integral of \( (x^2 - 2)^{11} x \): Let \( u = x^2 - 2 \), then \( du = 2x dx \). Rewrite integral: $$ \int (x^2 - 2)^{11} x \, dx = \frac{1}{2} \int u^{11} \, du = \frac{1}{2} \cdot \frac{u^{12}}{12} + C = \frac{(x^2 - 2)^{12}}{24} + C $$ ### Step 9 Integral of \( (5x^2 + 1)(5x^3 + 3x - 8)^6 \): Let \( u = 5x^3 + 3x - 8 \), then $$ du = (15x^2 + 3) dx = 3(5x^2 + 1) dx $$ Rewrite integral: $$ \int (5x^2 + 1)(5x^3 + 3x - 8)^6 \, dx = \frac{1}{3} \int u^6 \, du = \frac{1}{3} \cdot \frac{u^7}{7} + C = \frac{(5x^3 + 3x - 8)^7}{21} + C $$ --- **Final answers:** 1. \( \frac{x^4}{4} + C \) 2. \( \frac{7x^4}{4} + C \) 3. \( \frac{27x^8}{8} + \frac{x^6}{2} - 6x^4 + C \) 4. \( \frac{x^6}{6} + x^5 - \frac{3x^4}{4} + \frac{\sqrt{3} x^3}{3} + C \) 5. \( \frac{(10x^3 + 5)^7}{7} + C \) (assuming integrand includes \(30x^2\)) 6. \( \frac{(x^3 - 7)^9}{9} + C \) 7. \( \frac{(x^3 + 6x)^6}{6} + C \) 8. \( \frac{(x^2 - 2)^{12}}{24} + C \) 9. \( \frac{(5x^3 + 3x - 8)^7}{21} + C \)