Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 362

\[1)\ f(x) = \frac{2x^{4} - 4x^{3} + x}{3} =\]

\[= \frac{2}{3}x^{4} - \frac{4}{3}x^{3} + \frac{1}{3}x;\]

\[F(x) = \frac{2}{3} \bullet \frac{x^{5}}{5} - \frac{4}{3} \bullet \frac{x^{4}}{4} + \frac{1}{3} \bullet \frac{x^{2}}{2} + C =\]

\[= \frac{2}{15}x^{5} - \frac{1}{3}x^{4} + \frac{1}{6}x^{2} + C.\]

\[2)\ f(x) = \frac{6x^{3} - 3x + 2}{5} =\]

\[= \frac{6}{5}x^{3} - \frac{3}{5}x + \frac{2}{5};\]

\[F(x) = \frac{6}{5} \bullet \frac{x^{4}}{4} - \frac{3}{5} \bullet \frac{x^{2}}{2} + \frac{2}{5}x + C =\]

\[= \frac{3}{10}x^{4} - \frac{3}{10}x^{2} + \frac{2}{5}x + C.\]

\[3)\ f(x) = \frac{2x^{3} - 3x}{x^{2}} = 2x - \frac{3}{x};\]

\[F(x) = 2 \bullet \frac{x^{2}}{2} - 3\ln|x| + C =\]

\[= x^{2} - 3\ln|x| + C.\]

\[4)\ f(x) = \frac{3x^{4} + 5x^{2}}{x^{3}} = 3x + \frac{5}{x};\]

\[F(x) = 3 \bullet \frac{x^{2}}{2} + 5\ln|x| + C =\]

\[= \frac{3}{2}x^{2} + 5\ln|x| + C.\]

\[5)\ f(x) = 3x\left( 2 - x^{2} \right) = 6x - 3x^{3};\]

\[F(x) = 6 \bullet \frac{x^{2}}{2} - 3 \bullet \frac{x^{4}}{4} + C =\]

\[= 3x^{2} - \frac{3}{4}x^{4} + C.\]

\[6)\ f(x) = 2x(1 - x) = 2x - 2x^{2};\]

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

\[= x^{2} - \frac{2}{3}x^{3} + C.\]

\[7)\ f(x) = (1 + 2x)(x - 3) =\]

\[= 2x^{2} - 5x - 3;\]

\[F(x) = 2 \bullet \frac{x^{3}}{3} - 5 \bullet \frac{x^{2}}{2} - 3x + C =\]

\[= \frac{2}{3}x^{3} - \frac{5}{2}x^{2} - 3x + C.\]

\[8)\ f(x) = (2x - 3)(2 + 3x) =\]

\[= 6x^{2} - 5x - 6;\]

\[F(x) = 6 \bullet \frac{x^{3}}{3} - 5 \bullet \frac{x^{2}}{2} - 6x + C =\]

\[= 2x^{3} - \frac{5}{2}x^{2} - 6x + C.\]