ГДЗ по алгебре 11 класс Никольский Параграф 6. Первообразная и интеграл Задание 10

\[\boxed{\mathbf{10}.}\]

\[\textbf{а)}\ f(x) = \frac{1}{\sqrt{1 - x^{2}}}\]

\[\int_{}^{}{\frac{1}{\sqrt{1 - x^{2}}}dx = \arcsin x + C};\]

\[F(x) = \arcsin x + C.\]

\[\textbf{б)}\ f(x) = \frac{1}{1 + x^{2}};\]

\[\int_{}^{}{\frac{1}{1 + x^{2}}\text{dx}} = \arctan x + C;\]

\[F(x) = \arctan x + C.\]

\[\textbf{в)}\ f(x) = \frac{1}{\sqrt{1 - (2x)^{2}}};\]

\[\frac{1}{2}\int_{}^{}{\frac{1}{\sqrt{1 - (2x)^{2}}}d\left( 1 - (2x)^{2} \right)} =\]

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

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

\[\textbf{г)}\ f(x) = \frac{1}{1 + (3x)^{2}};\]

\[\frac{1}{3}\int_{}^{}{\frac{1}{1 + (3x)^{2}}d\left( 1 + (3x)^{2} \right)} =\]

\[= \frac{1}{3}\arctan{3x} + C;\]

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

\[\textbf{д)}\ f(x) = \frac{1}{\sqrt{1 - 9x^{2}}};\]

\[\frac{1}{3}\int_{}^{}{\frac{1}{\sqrt{1 - 9x^{2}}}d\left( 1 - 9x^{2} \right)} =\]

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

\[\textbf{е)}\ f(x) = \frac{1}{1 + 4x^{2}};\]

\[\frac{1}{2}\int_{}^{}{\frac{1}{1 + 4x^{2}}d\left( 1 + 4x^{2} \right)} =\]

\[= \frac{1}{2}\arctan{2x} + C;\]

\[F(x) = \frac{1}{2}\arctan{2x} + C.\]