Problem related to integration I tried using partial fractions but I am not sure whether it is the right approach or not. I need help with this problem
Evaluation of Integral $$\int\frac{3x^4-1}{(x^4+x+1)^2}dx $$

Let $$\displaystyle I = \int\frac{3x^4-1}{(x^4+x+1)^2}dx = \int\frac{3x^4-1}{x^2\left(x^3+1+x^{-1}\right)^2}dx = \int\frac{3x^2-x^{-2}}{(x^3+1+x^{-1})^2}dx$$
Now Put $x^3+1+x^{-1} = t\;,$ Then $(3x^2-x^{-2})dx = dt$
So Integral $$\displaystyle I = \int\frac{1}{t^2} dt = -\frac{1}{t}+\mathcal{C} = -\frac{1}{x^3+1+x^{-1}}+\mathcal{C}$$
So we get $$\displaystyle I = \int\frac{3x^4-1}{(x^4+x+1)^2}dx = -\left[\frac{x}{x^4+x+1}\right]+\mathcal{C}$$
