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!

Integrand can be transformed without long divison
$$\begin{gathered}\frac{x^6}{x^4 - 1}=\frac{x^6-1+1}{x^4 - 1}=\frac{(x^2-1)(x^4+x^2+1)}{x^4 - 1}+\frac{1}{x^4 - 1}=\\
=\frac{x^4+x^2+1}{x^2 + 1} + \frac{1}{x^4 - 1}=\\
=x^2+\frac{1}{x^2 + 1}+\frac{1}{2}\left(\frac{1}{x^2-1} - \frac{1}{x^2+1} \right)=\\
=x^2 + \frac{1}{2}\left(\frac{1}{x^2-1} + \frac{1}{x^2+1} \right).
\end{gathered}$$
