Решебник по алгебре 11 класс Никольский Параграф 6. Первообразная и интеграл Задание 17

\[\boxed{\mathbf{17}.}\]

\[\textbf{а)}\ \int_{}^{}\frac{\text{dx}}{\sqrt{1 - x^{2}}} = \arcsin x + C.\]

\[\textbf{б)}\ \int_{}^{}\frac{\text{dx}}{1 + x^{2}} = arctg\ x + C\]

\[\textbf{в)}\ \int_{}^{}\frac{\text{dx}}{\sqrt{1 - 2x^{2}}} =\]

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

\[\textbf{г)}\ \int_{}^{}\frac{\text{dx}}{1 + 3x^{2}} =\]

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

\[\textbf{д)}\ \int_{}^{}\frac{\text{dx}}{\sqrt{1 - (3x + 1)^{2}}} =\]

\[= \frac{1}{3}\arcsin{(3x + 1)} + C.\]

\[\textbf{е)}\ \int_{}^{}\frac{\text{dx}}{1 + (4x - 1)^{2}} =\]

\[= \frac{1}{4}\text{arctg\ }(4x - 1) + C.\]

\[\textbf{ж)}\ \int_{}^{}\frac{\text{dx}}{\sqrt{4x - 4x^{2}}} =\]

\[= \int_{}^{}\frac{\text{dx}}{\sqrt{1 - (2x - 1)^{2}}} =\]

\[= \frac{1}{2}\arcsin{(2x - 1)} + C.\]

\[\textbf{з)}\ \int_{}^{}\frac{\text{dx}}{4x^{2} + 12x + 10} =\]

\[= \int_{}^{}\frac{\text{dx}}{1 + (2x + 3)^{2}} =\]

\[= \frac{1}{2}\text{arctg\ }(2x + 3) + C.\]