ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 359

\[1)\ f(x) = \left( \frac{1}{2}x - 1 \right)^{7};\]

\[F(x) = 2 \bullet \frac{\left( \frac{1}{2}x - 1 \right)^{8}}{8} + C =\]

\[= \frac{1}{4}\left( \frac{1}{2}x - 1 \right)^{8} + C.\]

\[2)\ f(x) = \left( \frac{1}{3}x + 2 \right)^{5};\]

\[F(x) = 3 \bullet \frac{\left( \frac{1}{3}x + 2 \right)^{6}}{6} + C =\]

\[= \frac{1}{2}\left( \frac{1}{3}x + 2 \right)^{6} + C.\]

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

\[F(x) = \frac{1}{2} \bullet (2x - 3)^{\frac{7}{5}}\ :\frac{7}{5} =\]

\[= \frac{5}{14}(2x - 3)^{\frac{7}{5}} + C.\]

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

\[F(x) = \frac{1}{3} \bullet (3x - 1)^{\frac{7}{4}}\ :\frac{7}{4} =\]

\[= \frac{4}{21}(3x - 1)^{\frac{7}{4}} + C.\]

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

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

\[= \frac{9}{4}(2x - 1)^{\frac{2}{3}} + C.\]

\[6)\ f(x) = \frac{4}{\sqrt{4x + 1}} = 4(4x + 1)^{- \frac{1}{2}};\]

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

\[= 2\sqrt{4x + 1} + C.\]

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

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

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

\[8)\ f(x) = \sqrt[3]{2 - 3x} = (2 - 3x)^{\frac{1}{3}};\]

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

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