How to evaluate this integral: $\int \frac{x^4}{(x-1)(x^2-1)}dx\;$? $$\int \frac{x^4}{(x-1)(x^2-1)}dx$$
I tried to decompose the $(x^2-1)$ term into $(x+1)(x-1)$ thus getting $(x-1)^2(x+1)$ as the denominator. I can't use the method of partial fraction because of the $x^4$ term. Should I proceed through normal polynomial(I'm finding it difficult) division or is there any other methods using trigonometric substitution to solve this. Please provide only hints so that I can work it  out myself. 

Hint:
$$\frac{x^4}{(x-1)(x^2-1)}=\frac{x^4-1+1}{(x-1)(x^2-1)}=\frac{x^2+1}{x-1}+\frac{1}{(x-1)(x^2-1)}$$
$$=x+1+\frac{2}{x-1}+\frac{1}{2(x-1)}\left(\frac{1}{x-1}-\frac{1}{x+1}\right)$$
$$=x+1+\frac{2}{x-1}+\frac{1}{2(x-1)^2}-\frac{1}{4(x-1)}+\frac{1}{4(x+1)}$$
$$\to \int \frac{x^4}{(x-1)(x^2-1)}dx=\frac{x^2}{2}+x-\frac{1}{2(x-1)}+\frac{7}{4}\ln(x-1)+\frac{1}{4}\ln(x+1)+\text{C}$$
