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

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