Evaluating $\int \frac{1}{(x^4-1)^2}dx$ without partial fraction decomposition

Question

The substitution $x^2=\sec\theta$ doesn't seem to lead anywhere. I know the key is to manipulate it into a form $$\displaystyle \int \dfrac{\text{d}\left(x^a\pm\frac{1}{x^a}\right)}{f\left(x^a\pm\frac{1}{x^a}\right)}$$

But I fail to do so.

Answer 1

$$\int\dfrac{dx}{(x^4-1)^2}=\int\dfrac{x^3}{(x^4-1)^2}\cdot\dfrac1{x^3}dx$$

$$=\dfrac1{x^3}\cdot\int \dfrac{x^3}{(x^4-1)^2} dx+\dfrac34\int\dfrac{dx}{x^4(x^4-1)}$$

We can use $$\int\dfrac{dx}{x^4(x^4-1)}=\int\dfrac{x^4-(x^4-1)}{x^4(x^4-1)}dx$$

and $\displaystyle\int\dfrac2{x^4-1}=\int\dfrac{x^2+1-(x^2-1)}{x^4-1}=?$