Решение:
Используем правила интегрирования для нахождения первообразной.
- \( f(x) = x^6 \)
\( F(x) = \int x^6 dx = \frac{x^{6+1}}{6+1} + C = \frac{x^7}{7} + C \) - \( f(x) = x^3 - x^5 \)
\( F(x) = \int (x^3 - x^5) dx = \int x^3 dx - \int x^5 dx = \frac{x^4}{4} - \frac{x^6}{6} + C \) - \( f(x) = 4x^3 - 2x \)
\( F(x) = \int (4x^3 - 2x) dx = 4\int x^3 dx - 2\int x dx = 4\frac{x^4}{4} - 2\frac{x^2}{2} + C = x^4 - x^2 + C \) - \( f(x) = x - x^2 + 9 \)
\( F(x) = \int (x - x^2 + 9) dx = \int x dx - \int x^2 dx + \int 9 dx = \frac{x^2}{2} - \frac{x^3}{3} + 9x + C \) - \( f(x) = 5 - cos x \)
\( F(x) = \int (5 - cos x) dx = \int 5 dx - \int cos x dx = 5x - \sin x + C \) - \( f(x) = 6\sin x + 9 + 3x^6 \)
\( F(x) = \int (6\sin x + 9 + 3x^6) dx = 6\int \sin x dx + \int 9 dx + 3\int x^6 dx \)
\( F(x) = -6\cos x + 9x + 3\frac{x^7}{7} + C \) - \( f(x) = \frac{5}{\sqrt{x}} + x + 3 = 5x^{-\frac{1}{2}} + x + 3 \)
\( F(x) = \int (5x^{-\frac{1}{2}} + x + 3) dx = 5\int x^{-\frac{1}{2}} dx + \int x dx + \int 3 dx \)
\( F(x) = 5\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1} + \frac{x^2}{2} + 3x + C = 5\frac{x^{\frac{1}{2}}}{\frac{1}{2}} + \frac{x^2}{2} + 3x + C = 10\sqrt{x} + \frac{x^2}{2} + 3x + C \) - \( f(x) = 4\cos x + \frac{1}{7}x \)
\( F(x) = \int (4\cos x + \frac{1}{7}x) dx = 4\int \cos x dx + \frac{1}{7}\int x dx \)
\( F(x) = 4\sin x + \frac{1}{7}\frac{x^2}{2} + C = 4\sin x + \frac{x^2}{14} + C \) - \( f(x) = 2\sin x + 2x^2 - x - 3 \)
\( F(x) = \int (2\sin x + 2x^2 - x - 3) dx = 2\int \sin x dx + 2\int x^2 dx - \int x dx - \int 3 dx \)
\( F(x) = -2\cos x + 2\frac{x^3}{3} - \frac{x^2}{2} - 3x + C \)
Ответ: 1) \( \frac{x^7}{7} + C \); 2) \( \frac{x^4}{4} - \frac{x^6}{6} + C \); 3) \( x^4 - x^2 + C \); 4) \( \frac{x^2}{2} - \frac{x^3}{3} + 9x + C \); 5) \( 5x - \sin x + C \); 6) \( -6\cos x + 9x + \frac{3x^7}{7} + C \); 7) \( 10\sqrt{x} + \frac{x^2}{2} + 3x + C \); 8) \( 4\sin x + \frac{x^2}{14} + C \); 9) \( -2\cos x + \frac{2x^3}{3} - \frac{x^2}{2} - 3x + C \).