Решебник по алгебре 11 класс Никольский Параграф 6. Первообразная и интеграл Задание 8

\[\boxed{\mathbf{8}.}\]

\[\textbf{а)}\ f(x) = (5x - 20)^{20};\]

\[F(x) = \frac{1}{5} \cdot \frac{(5x - 2)^{21}}{21} + C =\]

\[= \frac{1}{105} \cdot (5x - 2)^{21} + C.\]

\[\textbf{б)}\ f(x) = \sqrt{x - 5};\]

\[F(x) = \frac{\left( \sqrt{x} - 5 \right)^{\frac{3}{2}}}{\frac{3}{2}} + C =\]

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

\[\textbf{в)}\ f(x) = 2^{x};\]

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

\[\textbf{г)}\ f(x) = 2^{3x - 1};\]

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

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

\[\textbf{д)}\ f(x) = 3^{x};\]

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

\[\textbf{е)}\ f(x) = 3^{2x + 7};\]

\[F(x) = \frac{1}{2} \cdot \frac{3^{2x + 7}}{\ln 3} + C =\]

\[= \frac{1}{2\ln 3} \cdot 3^{2x + 7} + C.\]

\[\textbf{ж)}\ f(x) = \frac{1}{4x - 2};\]

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

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

\[F(x) = - \frac{1}{5}\ln| - 5x + 2| + C.\]

\[\textbf{и)}\ f(x) = \frac{5}{x - 4};\]

\[F(x) = 5\ln{|x - 4|} + C.\]