Finding $ \int \frac{5x^2-x-4}{x^5+x^4+1}dx$ Finding $\displaystyle \int \frac{5x^2-x-4}{x^5+x^4+1}dx$
Attempt : $\displaystyle I = \int\frac{5x^2-x-4}{x^5+x^4+1}dx = \int\frac{5x^2-x-4}{(x^2+x+1)(x^3-x+1)}dx$
because $\omega,\omega$ are the roots of $x^5+x^4+1 = 0$
so one factor is $(x-\omega)(x-\omega^2) = (x^2+x+1)$
could some help me how to solve it, thanks

HINT:Using partial fraction decomposition
$$\frac{5x^2-x-4}{(x^2+x+1)(x^3-x+1)}=\frac{Ax+B}{x^2+x+1}+\frac{Cx^2+Dx+E}{x^3-x+1}$$
$${5x^2-x-4}=(Ax+B)(x^3-x+1)+(Cx^2+Dx+E)(x^2+x+1)$$
Solving gives..
$$\frac{5x^2-x-4}{(x^2+x+1)(x^3-x+1)}=\frac{-3x-3}{x^2+x+1}+\frac{3x^2-1}{x^3-x+1}$$
Another hint:
Maybe at some point you might require:$$\int\frac{1}{x^2+x+1}dx$$
Complete the square and use $$\int\frac{1}{a^2+x^2}dx=\frac{\arctan(\frac xa)}{a}+C$$$$$$
Can you do it now?
