Write out the Form of the Partial Fraction Decomposition: $\int\frac{x^6}{x^2-4}$ Write out the Form of the Partial Fraction Decomposition: $\int\frac{x^6}{x^2-4}$
The book says use long division my answer was $x^3+\frac{4x^3}{x^2-4}$
The answer manual is $\frac{x^6}{x^2-4}=x^4+4x^2+16+\frac{64}{(x+2)(x-2)}$

Notice, the formula $a^3-b^3=(a-b)(a^2+b^2+ab)$,
Now, re-arrange the numerator as follows  $$\frac{x^6}{x^2-4}=\frac{x^6-4^3+4^3}{x^2-4}$$
$$=\frac{((x^2)^3-4^3)+4^3}{x^2-4}$$
$$=\frac{(x^2)^3-4^3}{x^2-4}+\frac{64}{x^2-4}$$ $$=\frac{(x^2-4)((x^2)^2+4^2+4x^2)}{x^2-4}+\frac{64}{x^2-4}$$
$$=\frac{(x^2-4)(x^4+4x^2+16)}{x^2-4}+\frac{64}{x^2-4}$$
$$=x^4+4x^2+16+\frac{64}{x^2-4}$$
