Using the integral method by partial fractions result in the following integrals:${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{x^2+12x+12}{x^3-4x}dx$

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{2x^3-4x^2-15x+5}{x^2-2x-8}dx$

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{x^2}{x^4-2x^2-8}dx={\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{x^2}{(x+2)(x-2)(x^2+2)}dx$