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

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\[\textbf{а)}\ f(x) = \frac{11}{\text{co}s^{2}(3x + 11)};\ \]

\[F(x) = \frac{1}{3} \cdot tg\ (3x + 11) + C;\]

\[F^{'}(x) =\]

\[= \left( \frac{1}{3} \cdot tg\ (3x + 11) + C \right)^{'} =\]

\[= \left( \frac{1}{3} \cdot tg\ (3x + 11) \right)^{'} + C^{'} =\]

\[= \frac{1}{3}\left( \text{tg\ }(3x + 11) \right)^{'};\]

\[u_{x} = 3x + 11;\ \ y_{u} = \frac{1}{3}tg\ u;\]

\[F^{'}(x) = \frac{1}{3}\left( \text{tg\ u} \right)^{'} \cdot (3x + 11)^{'} =\]

\[= \frac{1}{3} \cdot \frac{1}{\text{co}s^{2}u} \cdot 3 = \frac{1}{\text{co}s^{2}u} =\]

\[= \frac{1}{\text{co}s^{2}(3x + 11)}.\]

\[F^{'}(x) = f(x) - значит,\ \]

\[F(x)\ первообразная\ для\ f(x).\]

\[\textbf{б)}\ f(x) = \frac{1}{\text{si}n^{2}( - 4x + 7)};\]

\[F(x) = \frac{1}{4}\text{ctg\ }( - 4x + 7) + C;\]

\[F^{'}(x) =\]

\[= \left( \frac{1}{4}\text{ctg\ }( - 4x + 7) + C \right)^{'} =\]

\[= \left( \frac{1}{4}\text{ctg\ }( - 4x + 7) \right)^{'} + C^{'} =\]

\[= \frac{1}{4}\left( \text{ctg\ }( - 4x + 7) \right)^{'};\]

\[u_{x} = - 4x + 7;\ \ y_{u} = \frac{1}{4}ctg\ u:\]

\[F^{'}(x) =\]

\[= \frac{1}{4}\left( \text{ctg\ u} \right)^{'} \cdot ( - 4x + 7)^{'} =\]

\[= \frac{1}{4} \cdot \left( - \frac{1}{\text{si}n^{2}u} \right) \cdot ( - 4) =\]

\[= \frac{1}{\text{si}n^{2}u} = \frac{1}{\text{si}n^{2}( - 4x + 7)}.\]

\[F^{'}(x) = f(x) - значит,\ \]

\[F(x)\ первообразная\ для\ f(x).\]

\[\textbf{в)}\ f(x) = e^{5x - 2} + e^{2x - 5};\]

\[F(x) = \frac{1}{5}e^{5x - 2} + \frac{1}{2}e^{2x - 5} + C;\]

\[F^{'}(x) =\]

\[= \left( \frac{1}{5}e^{5x - 2} + \frac{1}{2}e^{2x - 5} + C \right)^{'} =\]

\[= \left( \frac{1}{5}e^{5x - 2} \right)^{'} + \left( \frac{1}{2}e^{2x - 5} \right)^{'} + C^{'};\]

\[u_{x} = 5x - 2;\ \ y_{u} = \frac{1}{5}e^{u};\]

\[v_{x} = 2x - 5;\ \ y_{v} = \frac{1}{2}e^{v}:\]

\[= e^{u} + e^{v} = e^{5x - 2} + e^{2x - 5}.\]

\[F^{'}(x) = f(x) - значит,\ \]

\[F(x)\ первообразная\ для\ f(x).\]