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 $\frac{x^6}{x^2-4}=x^4+4x^2+16+\frac{64}{(x+2)(x-2)}$

The first step is dividing $x^6$ by $x^2$ to get $x^4$.
Then you subtract $x^6-x^4(x^2-4)=4x^4$ (as you also got). But you need to continue.
Second step is to divide $4x^4$ by $x^2$ to get $4x^2$.
Then you subtract $4x^4-4x^2(x^2-4)=16x^2$. Again you need to continue and dividing $16x^2$ by $x^2$ to get $16$.
Then you subtract $16x^2-16(x^2-4)=64$.
And your final answer is
$$\frac{x^6}{x^2-4}=x^4+4x^2+16+\frac{64}{x^2-4}.$$
