Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 988

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\[5)\ f(x) = 6x^{2} - 4x + 3 =\]

\[= 6x^{2} - 4x + 3x^{0};\]

\[F(x) = 6 \bullet \frac{x^{3}}{3} - 4 \bullet \frac{x^{2}}{2} + 3 \bullet \frac{x^{1}}{1} =\]

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

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

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

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

\[= 4 \bullet \frac{3}{4} \bullet x\sqrt[3]{x} - 6 \bullet \frac{2}{3} \bullet x\sqrt{x} =\]

\[= 3x\sqrt[3]{x} - 4x\sqrt{x} + C.\]