ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 990

\[\boxed{\mathbf{990}\mathbf{.}}\]

\[1)\ f(x) = (x + 1)^{4};\]

\[F(x) = \frac{(x + 1)^{5}}{1 \bullet 5} =\]

\[= \frac{1}{5}(x + 1)^{5} + C.\]

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

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

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

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

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

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

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

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

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

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

\[= \frac{9}{2} \bullet \sqrt[3]{(x + 3)^{2}} + C.\]

\[5)\ f(x) = \frac{1}{x - 1} + 4\cos(x + 2);\]

\[6)\ f(x) = \frac{3}{x - 3} - 2\sin(x - 1);\]

\[= 3\ln(x - 3) + 2\cos(x - 1) + C.\]