w8.5.2

$$ \int \frac{x^4}{(4-x^2)} dx

\implies -\int\frac{x^4}{(x^2-4)} dx

$$

by division

$$\displaystyle\int\dfrac{16}{x^2-4}+x^2+4 \ dx

\implies 16\int \frac{1}{(x+2)(x-2)}+ x^2+4 \ dx $$

$$\frac{1}{(x+2)(x-2)}=\frac{A}{x+2}+\frac{B}{x-2}

\displaystyle = -\dfrac{1}{4\left(x+2\right)}+\dfrac{1}{4\left(x-2\right)}

$$

so now we have

$$-\frac{1}{4}\int\dfrac{1}{x+2} dx

\ \ + \ \frac{1}{4}\int\dfrac{1}{x-2} dx

\ \ +\int {x}^{2} \ dx

+ \int 4 \ dx $$

just seeing if going down the right trail....