Самостоятельно: 1. ∫ x⁴dx; 2. f 4t³dt; 3. ∫(4u³ – 6u² - 4u + 3)du ; 4. ∫ 3(2x² - 1)²dx; 5. ** dx; 6. ∫(3x-4 + 8x-5)dx; dt 7. t2" 2dx 8.2 ; x+3 x² dx 9.∫x3+1 ;

Ответ:

1. \[\int x^4 dx = \frac{x^{4+1}}{4+1} + C = \frac{x^5}{5} + C\]

2. \[\int 4t^3 dt = 4 \int t^3 dt = 4 \cdot \frac{t^{3+1}}{3+1} + C = 4 \cdot \frac{t^4}{4} + C = t^4 + C\]

3. \[\int (4u^3 - 6u^2 - 4u + 3) du = 4 \int u^3 du - 6 \int u^2 du - 4 \int u du + 3 \int du = \]

   \[= 4 \cdot \frac{u^4}{4} - 6 \cdot \frac{u^3}{3} - 4 \cdot \frac{u^2}{2} + 3u + C = u^4 - 2u^3 - 2u^2 + 3u + C\]

4. \[\int 3(2x^2 - 1)^2 dx = 3 \int (4x^4 - 4x^2 + 1) dx = 3(4 \int x^4 dx - 4 \int x^2 dx + \int dx) = \]

   \[= 3(4 \cdot \frac{x^5}{5} - 4 \cdot \frac{x^3}{3} + x) + C = \frac{12x^5}{5} - 4x^3 + 3x + C\]

5. \[\int \frac{x^2 - x}{3x} dx = \frac{1}{3} \int \frac{x(x-1)}{x} dx = \frac{1}{3} \int (x - 1) dx = \frac{1}{3} (\int x dx - \int dx) = \]

   \[= \frac{1}{3} (\frac{x^2}{2} - x) + C = \frac{x^2}{6} - \frac{x}{3} + C\]

6. \[\int (3x^{-4} + 8x^{-5}) dx = 3 \int x^{-4} dx + 8 \int x^{-5} dx = 3 \cdot \frac{x^{-4+1}}{-4+1} + 8 \cdot \frac{x^{-5+1}}{-5+1} + C = \]

   \[= 3 \cdot \frac{x^{-3}}{-3} + 8 \cdot \frac{x^{-4}}{-4} + C = -x^{-3} - 2x^{-4} + C = -\frac{1}{x^3} - \frac{2}{x^4} + C\]

7. \[\int \frac{dt}{t^2} = \int t^{-2} dt = \frac{t^{-2+1}}{-2+1} + C = \frac{t^{-1}}{-1} + C = -\frac{1}{t} + C\]

8. \[\int \frac{2dx}{x+3} = 2 \int \frac{dx}{x+3} = 2 \ln|x+3| + C\]

9. \[\int \frac{x^2 dx}{x^3 + 1} = \frac{1}{3} \int \frac{3x^2 dx}{x^3 + 1} = \frac{1}{3} \ln|x^3 + 1| + C\]

Ответ: