Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 356

\[1)\ f(x) = 2x^{5} - 3x^{2};\]

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

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

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

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

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

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

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

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

\[4)\ f(x) = 6x^{2} - 4x + 3;\]

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

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

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

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

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

\[6)\ f(x) = 4\sqrt[3]{x} - 6\sqrt{x} =\]

\[= 4x^{\frac{1}{3}} - 6x^{\frac{1}{2}};\]

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

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

\[7)\ f(x) = \sqrt{x} + 2\sqrt[3]{x} = x^{\frac{1}{2}} + 2x^{\frac{1}{3}};\]

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

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

\[8)\ f(x) = \frac{2}{x^{3}} - \frac{3}{x} = 2x^{- 3} - \frac{3}{x};\]

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

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