Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 363

\[1)\ f(x) = (2x + 1)\sqrt{x} =\]

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

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

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

\[2)\ f(x) = (3x - 2)\sqrt[3]{x} =\]

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

\[F(x) = 3 \bullet x^{\frac{7}{3}}\ :\frac{7}{3} - 2 \bullet x^{\frac{4}{3}}\ :\frac{4}{3} + C =\]

\[= \frac{9}{7}x^{\frac{7}{3}} - \frac{3}{2}x^{\frac{4}{3}} + C.\]

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

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

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

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

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

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

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

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