Решебник по алгебре 11 класс Никольский Параграф 6. Первообразная и интеграл Задание 22

\[\boxed{\mathbf{22}.}\]

\[\textbf{а)}\ \int_{}^{}\frac{4xdx}{1 + x^{2}}\]

\[1 + x^{2} = t\]

\[2xdx = dt\]

\[xdx = \frac{1}{2}dt:\]

\[\int_{}^{}\frac{4 \cdot \frac{1}{2}\text{dt}}{t} = \int_{}^{}\frac{2dt}{t} = 2\int_{}^{}\frac{\text{dt}}{t} =\]

\[= 2 \cdot \ln|t| + C =\]

\[= 2\ln\left| 1 + x^{2} \right| + C =\]

\[= 2\ln\left( 1 + x^{2} \right) + C;\ \ \]

\[так\ как\ \left( 1 + x^{2} \right) > 0.\]

\[\textbf{б)}\ \int_{}^{}\frac{2xdx}{1 + 4x^{2}}\]

\[1 + 4x^{2} = t\]

\[8xdx = dt\]

\[xdx = \frac{1}{8}dt:\]

\[\int_{}^{}\frac{2 \cdot \frac{1}{8}\text{dt}}{t} = \int_{}^{}\frac{\frac{1}{4}\text{dt}}{t} = \frac{1}{4}\int_{}^{}\frac{\text{dt}}{t} =\]

\[= \frac{1}{4} \cdot \ln|t| + C =\]

\[= \frac{1}{4}\ln\left| 1 + 4x^{2} \right| + C =\]

\[= \frac{1}{4}\ln\left( 1 + {4x}^{2} \right) + C;\ \ \]

\[так\ как\ \left( 1 + {4x}^{2} \right) > 0.\]

\[\textbf{в)}\ \int_{}^{}\frac{\text{xdx}}{1 + 9x^{2}}\]

\[1 + 9x^{2} = t\]

\[18xdx = dt\]

\[xdx = \frac{1}{18}dt:\]

\[\int_{}^{}\frac{\frac{1}{18}\text{dt}}{t} = \frac{1}{18}\int_{}^{}\frac{\text{dt}}{t} =\]

\[= \frac{1}{18} \cdot \ln|t| + C =\]

\[= \frac{1}{18}\ln\left| 1 + 9x^{2} \right| + C =\]

\[= \frac{1}{18}\ln\left( 1 + 9x^{2} \right) + C;\ \ \]

\[так\ как\ \left( 1 + 9x^{2} \right) > 0.\]

\[\textbf{г)}\ \int_{}^{}\frac{\text{xdx}}{4 + x^{2}}\]

\[4 + x^{2} = t\]

\[2xdx = dt\]

\[xdx = \frac{1}{2}dt:\]

\[\int_{}^{}\frac{\frac{1}{2}\text{dt}}{t} = \frac{1}{2}\int_{}^{}\frac{\text{dt}}{t} = \frac{1}{2} \cdot \ln|t| + C =\]

\[= \frac{1}{2}\ln\left| 4 + x^{2} \right| + C =\]

\[= \frac{1}{2}\ln\left( 4 + x^{2} \right) + C;\ \ \]

\[так\ как\ \left( 4 + x^{2} \right) > 0.\]