I'm having a lot of trouble with this integral. I don't know what to set A+B equal to. There's an x^4 in the numerator and I'm trying to figure out how to account for it. What am I doing wrong? Thanks.
Problem: $$\int\frac{x^4}{4-x^2}\space dx$$
What I have tried:
$$\frac{x^4}{(2-x)(2+x)}=\frac{a}{2-x}+\frac{b}{2+x}=\frac{a(2+x)+b(2-x)}{(2-x)(2+x)}\implies x^4=(a+b)x+2(a+b)$$
When the numerator has higher degree, you should do long division before partial fractions.
$$\frac{x^4}{4-x^2} = -x^2 - 4 - \frac{16}{x^2-4}.$$ Now you can do partial fractions on the last term.