How to integrate $\frac{4x+4}{x^4+x^3+2x^2}$? Please could anyone help me to integrate
$\quad\displaystyle{4x + 4 \over x^4 + x^3 + 2x^2}.\quad$
I know how to use partial fraction and I did this: 
$$
x^{4} + x^{3} + 2x^{2} = x^{2}\left(x^{2} + x + 2\right)
$$
And then ?.$\quad$ Thanks all.

We have:
$$ \int{\frac{4x+4}{x^4+x^3+2x^2}} dx $$
We need to decompose this fraction into pieces and then integrate each one separately. We start by simplifying the denominator:
$$ \frac{4x+4}{x^2(x^2+x+2)} $$
Since we have two quadratic terms in the denominator, we can guess the form:
$$ \frac{4x+4}{x^2(x^2+x+2)} = \frac{Ax+B}{x^2} + \frac{Cx+D}{x^2+x+2} $$
Simplifying this further yields:
$$ 4x+4 = (Ax+B)(x^2+x+2) + x^2(Cx+D) $$
Let's expand everything out so that we can solve for these constants A, B, C, and D:
$$ 4x+4 = (Ax^3+Ax^2+2Ax+Bx^2+Bx+2B)+(Cx^3+Dx^2) $$
$$ 4x+4 = (A+C)x^3+(A+B+D)x^2+(2A+B)x+2B $$
We now have a system of equations: 
$$ A+C=0, A+B+D=0, 2A+B=4, 2B=4 $$
Solving (by working with the 2B=4 term first), we get:
$$ A=1, B=2, C=-1, D=-3$$
Plugging back our constants into our "guess form":
$$\frac{4x+4}{x^2(x^2+x+2)}=\frac{(1)x+(2)}{x^2}+\frac{(-1)x+(-3)}{x^2+x+2}$$
$$=\frac{x+2}{x^2} + \frac{-x-3}{x^2+x+2}$$
$$=\frac{1}{x}+\frac{2}{x^2}-\frac{x-3}{x^2+x+2} $$
That's much better, now you should be able to integrate each piece separately.
Can you finish?
