<s>[INST] How to evaluate $\int \frac{x^6}{x^4-1} \, \mathrm{d}x.$ Evaluate the integral:
$$\int \frac{x^6}{x^4-1} \, \mathrm{d}x$$
After a lot of help I have reached this point:
$x^2 = Ax^3 - Ax + Bx^2 - B + Cx^3 + Cx^2 + Cx + C + Dx^3 - Dx^2 + Dx - D$
But now I don't really know how to solve for $A, B, C$, and $D$. Please help!
 [/INST] First: Divide! Use polynomial division to get $$\frac{x^6}{x^4 - 1} = 1 + \frac{x^2}{x^4 - 1}$$ $\int 1\,dx = x + C$
For the second term:
Now factor the denominator, and decompose: $$x^4 - 1 = (x^2 + 1)(x^2-1) = (x^2 + 1)(x-1)(x+1)$$
So the set up we want for the second term is:
$$\frac{x^2}{x^4 - 1} = \dfrac{Ax + B}{x^2 + 1} +\frac{C}{x-1} + \frac D{x+1}$$
Now solve for A, B, C, D.
</s>