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

Your long division is wrong and you can easily check it by doing the computation:
$$
x^3+\frac{4x^3}{x^2-4}=\frac{x^5}{x^2-4}
$$
Moreover, the degree of the remainder should be less than the degree of the denominator.
If the long division is performed correctly, you find
$$
\frac{x^6}{x^2-4}=x^4+4x^2+16+\frac{64}{x^2-4}
$$
but this is not the end of the job, because you must also split the last fraction as
$$
\frac{64}{x^2-4}=\frac{A}{x-2}+\frac{B}{x+2}
$$
that entails
\begin{cases}
A+B=0\\[6px]
2A-2B=64
\end{cases}
that means $A=16$, $B=-16$. Thus the final decomposition is
$$
\frac{x^6}{x^2-4}=x^4+4x^2+16+\frac{16}{x-2}-\frac{16}{x+2}
$$
from which the integral can be easily computed.

Here's the long division:

