∫ ( x 2 + x − 2 ) d x = x 3 3 + x − 1 − 1 + C = 1 3 x 3 − 1 x + C {\displaystyle \int _{}^{}\left(x^{2}+x^{-2}\right)dx={\frac {x^{3}}{3}}+{\frac {x^{-}1}{-1}}+C={\frac {1}{3}}x^{3}-{\frac {1}{x}}+C}
