ГДЗ по алгебре 11 класс Никольский Параграф 6. Первообразная и интеграл Задание 19

\[\boxed{\mathbf{19}.}\]

\[\textbf{а)}\ \int_{}^{}{e^{3x}\text{dx}};\]

\[3x = t;\ \ 3dx = dt;\ \ dx = \frac{1}{3}dt:\]

\[\int_{}^{}{e^{t} \cdot \frac{1}{3}\text{dt}} = \frac{1}{3}\int_{}^{}{e^{t}\text{dt}} =\]

\[= \frac{1}{3}e^{t} + C = \frac{1}{3}e^{3x} + C.\]

\[\textbf{б)}\ \int_{}^{}{9^{2x}\text{dx}};\]

\[2x = t;\ \ 2dx = dt;dx = \frac{1}{2}dt:\]

\[\int_{}^{}{9^{t} \cdot \frac{1}{2}\text{dt}} = \frac{1}{2}\int_{}^{}{9^{t} \cdot dt} =\]

\[= \frac{1}{2} \cdot \frac{9^{t}}{\ln 9} + C = \frac{9^{t}}{2\ln 9} + C =\]

\[= \frac{9^{2x}}{2\ln 9} + C.\]

\[\textbf{в)}\ \int_{}^{}{\sin{7x}\text{dx}};\]

\[7x = t;\ \ 7dx = dt;\ \ dx = \frac{1}{7}dt:\]

\[\int_{}^{}{\sin t \cdot \frac{1}{7}\text{dt}} = \frac{1}{7}\int_{}^{}{\sin t\text{dt}} =\]

\[= \frac{1}{7} \cdot \left( - \cos t \right) + C =\]

\[= - \frac{\cos t}{7} + C = - \frac{\cos{7x}}{7} + C.\]

\[\textbf{г)}\ \int_{}^{}{\cos{4x}\text{dt}};\]

\[4x = t;\ \ 4dx = dt;\ \ dx = \frac{1}{4}dt:\]

\[\int_{}^{}{\cos t \cdot \frac{1}{4}\text{dt}} = \frac{1}{4}\int_{}^{}{\cos t\text{dt}} =\]

\[= \frac{1}{4} \cdot \sin t + C = \frac{\sin{4x}}{4} + C.\]

\[\textbf{д)}\ \int_{}^{}{\sqrt{7 - 2x}\text{dx}};\]

\[7 - 2x = t;\]

\[7dx = dt;\ \ dx = \frac{1}{7}dt:\]

\[\int_{}^{}{\sqrt{t} \cdot \frac{1}{7}\text{dt}} = \frac{1}{7}\int_{}^{}{t^{\frac{1}{2}}\text{dt}} =\]

\[= \frac{1}{7} \cdot \frac{t^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} + C = \frac{t^{\frac{3}{2}}}{7 \cdot \frac{3}{2}} + C =\]

\[= \frac{2t^{\frac{3}{2}}}{21} + C =\]

\[= \frac{2(7x - 2)^{\frac{3}{2}}}{21} + C =\]

\[= \frac{2\sqrt{(7x - 2)^{3}}}{21} + C.\]

\[\textbf{е)}\ \int_{}^{}{\sqrt[3]{(2x + 1)^{2}}\text{dx}};\]

\[2x + 1 = t;\]

\[2dx = dt;\ \ dx = \frac{1}{2}dt:\]

\[\int_{}^{}{\sqrt[3]{t^{2}} \cdot \frac{1}{2}\text{dt}} = \frac{1}{2}\int_{}^{}{t^{\frac{2}{3}}\text{\ dt}} =\]

\[= \frac{1}{2} \cdot \frac{t^{\frac{2}{3} + 1}}{\frac{2}{3} + 1} + C = \frac{t^{\frac{3}{5}}}{2 \cdot \frac{5}{3}} + C =\]

\[= \frac{3t^{\frac{5}{3}}}{10} + C = \frac{3(2x + 1)^{\frac{5}{3}}}{10} + C =\]

\[= \frac{3\sqrt[3]{(2x + 1)^{5}}}{10} + C.\]