- #1
karush
Gold Member
MHB
- 3,269
- 5
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...
$$ \int \frac{x^4}{(4-x^2)} dx
\implies -\int\frac{x^4}{(x^2-4)} dx
$$
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...
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