ГДЗ по алгебре 11 класс Никольский Параграф 6. Первообразная и интеграл Задание 20

\[\boxed{\mathbf{20}.}\]

\[\textbf{а)}\ \int_{}^{}\frac{\text{xdx}}{\sqrt{4 - x^{2}}};\]

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

\[- 2xdx = dt\]

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

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

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

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

\[= - \sqrt{4 - x^{2}} + C.\]

\[\textbf{б)}\ \int_{}^{}\frac{3xdx}{\sqrt{25 - x^{2}}};\]

\[25 - x^{2} = t\]

\[- 2xdx = dt\]

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

\[\int_{}^{}\frac{3 \cdot \left( - \frac{1}{2} \right)\text{dt}}{\sqrt{t}} = - \frac{3}{2}\int_{}^{}{t^{- \frac{1}{2}} \cdot dt} =\]

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

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

\[= - 3\sqrt{25 - x^{2}} + C.\]

\[\textbf{в)}\ \int_{}^{}\frac{2xdx}{\sqrt{9 - 4x^{2}}};\]

\[9 - 4x^{2} = t\]

\[- 8xdx = dt\]

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

\[\int_{}^{}\frac{2 \cdot \left( - \frac{1}{8} \right)\text{dt}}{\sqrt{t}} = - \frac{1}{4}\int_{}^{}{t^{- \frac{1}{2}} \cdot dt} =\]

\[= - \frac{1}{4} \cdot \frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} + C =\]

\[= - \frac{t^{\frac{1}{2}}}{4 \cdot \frac{1}{2}} + C = - \frac{\sqrt{t}}{2} + C =\]

\[= - \frac{1}{2}\sqrt{9 - 4x^{2}} + C.\]

\[\textbf{г)}\ \int_{}^{}\frac{5xdx}{\sqrt{4 - 9x^{2}}};\]

\[4 - 9x^{2} = t\]

\[- 18xdx = dt\]

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

\[\int_{}^{}\frac{5 \cdot \left( - \frac{1}{18} \right)\text{dt}}{\sqrt{t}} =\]

\[= - \frac{5}{18}\int_{}^{}{t^{- \frac{1}{2}} \cdot dt} =\]

\[= - \frac{5}{18} \cdot \frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} + C =\]

\[= - \frac{5}{18} \cdot \frac{t^{\frac{1}{2}}}{\frac{1}{2}} + C = - \frac{5}{9}\sqrt{t} + C =\]

\[= - \frac{5}{9}\sqrt{4 - 9x^{2}} + C.\]