How to integrate $\frac{4x}{x^3+x^2+x+1}$ I'd like to know how to integrate $$\frac{4x}{x^3+x^2+x+1}$$
Please could anyone help me 
Thanks all

Note that $x^3+x^2+x+1=(x+1)(x^2+1)$. We use partial fracions. So we try to find constants $A,B,C$ such that
$$\frac{4x}{(x+1)(x^2+1)}=\frac{A}{x+1}+\frac{Bx+C}{x^2+1}.$$
Bring the right-hand side to the common denominator $(x+1)(x^2+1)$. The numerators must be identically equal. It follows that
$$4x=A(x^2+1)+(Bx+C)(x+1).$$
Set $x=-1$. We get $-4=2A$, and therefore $A=-2$.
On the right, the coefficient of $x^2$ is $-2+B$. On the left it is $0$. It follows that $B=2$.
On the right, the coefficient of $x$ is therefore $2+C$. Thus $C=2$. We conclude that
$$\frac{4x}{(x+1)(x^2+1)}=-\frac{2}{x+1}+\frac{2x+2}{x^2+1}.$$
So we want to integrate $-\frac{2}{x+1}+\frac{2x}{x^2+1}+\frac{2}{x^2+1}$.  
The rest is straightforward. To integrate $\frac{2x}{x^2+1}$ let $u=x^2+1$. 
