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