Предмет: Математика, автор: НиколайСПБ

Найти неопределенный интеграл

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Автор ответа: NNNLLL54
1
\int \Big (\sqrt[5]{(1-x)^2}+sin3x-\frac{1}{9-x^2}-e^{-x}\Big )dx=\\\\=-\int (1-x)^{2/5}d(1-x)+\frac{1}{3}\int sin3x\cdot d(3x)+\int \frac{dx}{x^2-3^2}+\\\\+\int e^{-x}\cdot d(-x)=-\frac{5(1-x)^{7/5}}{7}-\frac{1}{3}\cdot cos3x+\frac{1}{2\cdot 3}\cdot ln\Big |\frac{x-3}{x+3}\Big |+e^{-x}+C

НиколайСПБ: Я даже и не знаю какой ответ выставить лучшим)
Автор ответа: Alexаndr
3
\int(\sqrt[5]{(1-x)^2}+sin3x-\frac{1}{9-x^2}-e^{-x})=-\int\sqrt[5]{(1-x)^2}d(1-x)+\\+\frac{1}{3}\int sin3xd(3x)+\int\frac{dx}{x^2-9}+\int e^{-x}d(-x)=-\frac{5\sqrt[5]{(1-x)^7}}{7}-\frac{cos3x}{3}+\\+\frac{1}{6}ln|\frac{x-3}{x+3}|+e^{-x}+C
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