1. **Problem:** Evaluate the integral $$\int \sqrt{x - 2} \, dx$$ 2. **Formula and substitution:** Use substitution $u = x - 2$, so $du = dx$. The integral becomes $$\int u^{1/2} \, du$$ 3. **Integration:** Use the power rule for integration $$\int u^n \, du = \frac{u^{n+1}}{n+1} + C$$ for $n \neq -1$. 4. **Apply the rule:** $$\int u^{1/2} \, du = \frac{u^{3/2}}{3/2} + C = \frac{2}{3} u^{3/2} + C$$ 5. **Back-substitute:** Replace $u$ with $x - 2$ to get $$\frac{2}{3} (x - 2)^{3/2} + C$$ --- 1. **Problem:** Evaluate the integral $$\int (2x + 3)^{11} \, dx$$ 2. **Substitution:** Let $u = 2x + 3$, then $du = 2 dx$ or $dx = \frac{1}{2} du$. 3. **Rewrite integral:** $$\int (2x + 3)^{11} \, dx = \int u^{11} \cdot \frac{1}{2} du = \frac{1}{2} \int u^{11} \, du$$ 4. **Integrate:** $$\frac{1}{2} \cdot \frac{u^{12}}{12} + C = \frac{u^{12}}{24} + C$$ 5. **Back-substitute:** $$\frac{(2x + 3)^{12}}{24} + C$$ --- 1. **Problem:** Evaluate $$\int \sqrt{5x - 1} \, dx$$ 2. **Substitution:** Let $u = 5x - 1$, so $du = 5 dx$ or $dx = \frac{1}{5} du$. 3. **Rewrite integral:** $$\int \sqrt{5x - 1} \, dx = \int u^{1/2} \cdot \frac{1}{5} du = \frac{1}{5} \int u^{1/2} \, du$$ 4. **Integrate:** $$\frac{1}{5} \cdot \frac{u^{3/2}}{3/2} + C = \frac{1}{5} \cdot \frac{2}{3} u^{3/2} + C = \frac{2}{15} u^{3/2} + C$$ 5. **Back-substitute:** $$\frac{2}{15} (5x - 1)^{3/2} + C$$ --- 1. **Problem:** Evaluate $$\int \sqrt{6x + 1} \, dx$$ 2. **Substitution:** Let $u = 6x + 1$, so $du = 6 dx$ or $dx = \frac{1}{6} du$. 3. **Rewrite integral:** $$\int (6x + 1)^{1/2} \, dx = \int u^{1/2} \cdot \frac{1}{6} du = \frac{1}{6} \int u^{1/2} \, du$$ 4. **Integrate:** $$\frac{1}{6} \cdot \frac{u^{3/2}}{3/2} + C = \frac{1}{6} \cdot \frac{2}{3} u^{3/2} + C = \frac{1}{9} u^{3/2} + C$$ 5. **Back-substitute:** $$\frac{1}{9} (6x + 1)^{3/2} + C$$ --- 1. **Problem:** Evaluate $$\int 5 (3 - 4x)^{2/3} \, dx$$ 2. **Substitution:** Let $u = 3 - 4x$, so $du = -4 dx$ or $dx = -\frac{1}{4} du$. 3. **Rewrite integral:** $$\int 5 u^{2/3} \, dx = 5 \int u^{2/3} \cdot -\frac{1}{4} du = -\frac{5}{4} \int u^{2/3} \, du$$ 4. **Integrate:** $$-\frac{5}{4} \cdot \frac{u^{5/3}}{5/3} + C = -\frac{5}{4} \cdot \frac{3}{5} u^{5/3} + C = -\frac{3}{4} u^{5/3} + C$$ 5. **Back-substitute:** $$-\frac{3}{4} (3 - 4x)^{5/3} + C$$ --- 1. **Problem:** Evaluate $$\int \frac{dx}{(8x - 1)^3}$$ 2. **Substitution:** Let $u = 8x - 1$, so $du = 8 dx$ or $dx = \frac{1}{8} du$. 3. **Rewrite integral:** $$\int \frac{dx}{u^3} = \int u^{-3} \cdot \frac{1}{8} du = \frac{1}{8} \int u^{-3} \, du$$ 4. **Integrate:** $$\frac{1}{8} \cdot \frac{u^{-2}}{-2} + C = -\frac{1}{16} u^{-2} + C = -\frac{1}{16 (8x - 1)^2} + C$$ --- 1. **Problem:** Evaluate $$\int x (x^2 + 2)^6 \, dx$$ 2. **Substitution:** Let $u = x^2 + 2$, so $du = 2x dx$ or $x dx = \frac{1}{2} du$. 3. **Rewrite integral:** $$\int x (x^2 + 2)^6 \, dx = \int u^6 \cdot x dx = \int u^6 \cdot \frac{1}{2} du = \frac{1}{2} \int u^6 \, du$$ 4. **Integrate:** $$\frac{1}{2} \cdot \frac{u^7}{7} + C = \frac{u^7}{14} + C$$ 5. **Back-substitute:** $$\frac{(x^2 + 2)^7}{14} + C$$ --- 1. **Problem:** Evaluate $$\int 6x^2 \sqrt{3x^3 - 1} \, dx$$ 2. **Substitution:** Let $u = 3x^3 - 1$, so $du = 9x^2 dx$ or $x^2 dx = \frac{1}{9} du$. 3. **Rewrite integral:** $$\int 6x^2 u^{1/2} \, dx = 6 \int u^{1/2} x^2 dx = 6 \int u^{1/2} \cdot \frac{1}{9} du = \frac{2}{3} \int u^{1/2} \, du$$ 4. **Integrate:** $$\frac{2}{3} \cdot \frac{u^{3/2}}{3/2} + C = \frac{2}{3} \cdot \frac{2}{3} u^{3/2} + C = \frac{4}{9} u^{3/2} + C$$ 5. **Back-substitute:** $$\frac{4}{9} (3x^3 - 1)^{3/2} + C$$ --- 1. **Problem:** Evaluate $$\int (1 + \frac{1}{x})^3 \cdot \frac{1}{x^2} \, dx$$ 2. **Substitution:** Let $u = 1 + \frac{1}{x}$, then $$du = -\frac{1}{x^2} dx \Rightarrow -du = \frac{1}{x^2} dx$$ 3. **Rewrite integral:** $$\int u^3 \cdot \frac{1}{x^2} dx = \int u^3 (-du) = -\int u^3 \, du$$ 4. **Integrate:** $$- \frac{u^4}{4} + C = -\frac{(1 + \frac{1}{x})^4}{4} + C$$ --- 1. **Problem:** Evaluate $$\int x^{1/3} (x^{1/3} + 9)^8 \, dx$$ 2. **Substitution:** Let $u = x^{1/3} + 9$, so $du = \frac{1}{3} x^{-2/3} dx$ or $dx = 3 x^{2/3} du$. 3. **Rewrite integral:** Express $x^{1/3}$ and $dx$ in terms of $u$ and $du$: Since $u = x^{1/3} + 9$, then $x^{1/3} = u - 9$. Rewrite integral: $$\int x^{1/3} (x^{1/3} + 9)^8 dx = \int (u - 9) u^8 dx$$ But $dx = 3 x^{2/3} du = 3 (x^{1/3})^2 du = 3 (u - 9)^2 du$ So integral becomes: $$\int (u - 9) u^8 \cdot 3 (u - 9)^2 du = 3 \int u^8 (u - 9)^3 du$$ 4. **Expand:** $$(u - 9)^3 = u^3 - 27 u^2 + 243 u - 729$$ So integral is: $$3 \int u^8 (u^3 - 27 u^2 + 243 u - 729) du = 3 \int (u^{11} - 27 u^{10} + 243 u^9 - 729 u^8) du$$ 5. **Integrate term-by-term:** $$3 \left( \frac{u^{12}}{12} - 27 \frac{u^{11}}{11} + 243 \frac{u^{10}}{10} - 729 \frac{u^9}{9} \right) + C$$ 6. **Simplify coefficients:** $$= \frac{3}{12} u^{12} - \frac{81}{11} u^{11} + \frac{729}{10} u^{10} - 243 u^9 + C$$ 7. **Back-substitute:** $$= \frac{1}{4} (x^{1/3} + 9)^{12} - \frac{81}{11} (x^{1/3} + 9)^{11} + \frac{729}{10} (x^{1/3} + 9)^{10} - 243 (x^{1/3} + 9)^9 + C$$ --- 1. **Problem:** Evaluate $$\frac{2}{3} \int \sqrt{4 - \frac{3}{5} x} \, dx$$ 2. **Substitution:** Let $u = 4 - \frac{3}{5} x$, so $du = -\frac{3}{5} dx$ or $dx = -\frac{5}{3} du$. 3. **Rewrite integral:** $$\frac{2}{3} \int u^{1/2} \cdot -\frac{5}{3} du = -\frac{10}{9} \int u^{1/2} \, du$$ 4. **Integrate:** $$-\frac{10}{9} \cdot \frac{u^{3/2}}{3/2} + C = -\frac{10}{9} \cdot \frac{2}{3} u^{3/2} + C = -\frac{20}{27} u^{3/2} + C$$ 5. **Back-substitute:** $$-\frac{20}{27} \left(4 - \frac{3}{5} x \right)^{3/2} + C$$ --- 1. **Problem:** Evaluate $$\int (3x + 15) \sqrt{x^2 + 10x + 4} \, dx$$ 2. **Rewrite:** Note $3x + 15 = 3(x + 5)$ and complete the square inside the root: $$x^2 + 10x + 4 = (x + 5)^2 - 21$$ 3. **Substitution:** Let $u = x + 5$, so $du = dx$. 4. **Rewrite integral:** $$\int 3u \sqrt{u^2 - 21} \, du = 3 \int u \sqrt{u^2 - 21} \, du$$ 5. **Substitution:** Let $v = u^2 - 21$, so $dv = 2u du$ or $u du = \frac{1}{2} dv$. 6. **Rewrite integral:** $$3 \int u \sqrt{v} \, du = 3 \int \sqrt{v} \cdot \frac{1}{2} dv = \frac{3}{2} \int v^{1/2} \, dv$$ 7. **Integrate:** $$\frac{3}{2} \cdot \frac{v^{3/2}}{3/2} + C = v^{3/2} + C$$ 8. **Back-substitute:** $$\left(u^2 - 21\right)^{3/2} + C = \left((x + 5)^2 - 21\right)^{3/2} + C$$