Решебник по алгебре 11 класс Никольский Параграф 6. Первообразная и интеграл Задание 21

\[\boxed{\mathbf{21}.}\]

\[\textbf{а)}\ \int_{}^{}{x\sqrt{1 + x^{2}}\text{dx}}\]

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

\[2xdx = dt\]

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

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

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

\[= \frac{1}{3}\sqrt{t^{3}} + C =\]

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

\[\textbf{б)}\ \int_{}^{}{5x\sqrt{1 + 4x^{2}}\text{dx}}\]

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

\[8xdx = dt\]

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

\[5\int_{}^{}{\sqrt{t} \cdot \frac{1}{8}\text{dt}} = 5 \cdot \frac{1}{8}\int_{}^{}{t^{\frac{1}{2}} \cdot dt} =\]

\[= \frac{5}{8} \cdot \frac{t^{\frac{3}{2}}}{\frac{3}{2}} = \frac{5}{8} \cdot \frac{2}{3} \cdot t^{\frac{3}{2}} + C =\]

\[= \frac{5}{12}\sqrt{t^{3}} + C =\]

\[= \frac{5}{12}\sqrt{\left( 1 + 4x^{2} \right)^{3}} + C.\]

\[\textbf{в)}\ \int_{}^{}{x\sqrt{4 + x^{2}}\text{dx}}\]

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

\[2xdx = dt\]

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

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

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

\[= \frac{1}{3}\sqrt{t^{3}} + C =\]

\[= \frac{1}{3}\sqrt{\left( 4 + x^{2} \right)^{3}} + C.\]

\[\textbf{г)}\ \int_{}^{}{x\sqrt{9 + x^{2}}\text{dx}}\]

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

\[2xdx = dt\]

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

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

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

\[= \frac{1}{3}\sqrt{t^{3}} + C =\]

\[= \frac{1}{3}\sqrt{\left( 9 + x^{2} \right)^{3}} + C\]