ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 991

\[\boxed{\mathbf{991}\mathbf{.}}\]

\[1)\ f(x) = \sin(2x + 3);\]

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

\[2)\ f(x) = \cos(3x + 4);\]

\[F(x) = \frac{1}{3}\sin(3x + 4) + C.\]

\[3)\ f(x) = \cos\left( \frac{x}{2} - 1 \right);\]

\[F(x) = 1\ :\frac{1}{2} \bullet \sin\left( \frac{x}{2} - 1 \right) =\]

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

\[4)\ f(x) = \sin\left( \frac{x}{4} + 5 \right);\]

\[F(x) = 1\ :\left( - \frac{1}{4} \right) \bullet \cos\left( \frac{x}{4} + 5 \right) =\]

\[= - 4\cos\left( \frac{x}{4} + 5 \right) + C.\]

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

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

\[6)\ f(x) = e^{3x - 5};\]

\[F(x) = \frac{1}{3}e^{3x - 5} + C.\]

\[7)\ f(x) = \frac{1}{2x} = \frac{1}{2} \bullet \frac{1}{x};\]

\[F(x) = \frac{1}{2} \bullet \ln x + C.\]

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

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