Question

Решите пж Интеграл :)

Answer 1

$\displaystyle\int\frac{\sqrt[7]{2+\displaystyle\frac{1}{x^2}}}{x^3}dx=[\frac{dx}{x^3}=-\frac{1}{2}d(2+\frac{1}{x^2})]=\\=-\frac{1}{2}\int\sqrt[7]{2+\displaystyle\frac{1}{x^2}}d(2+\displaystyle\frac{1}{x^2})=-\frac{7}{16}\sqrt[7]{(2+\displaystyle\frac{1}{x^2})^8}+C$