I'd like to know how to integrate $$\frac{4x}{x^3+x^2+x+1}$$
Please could anyone help me
Thanks all
$\endgroup$ 62 Answers
$\begingroup$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 $u=x^2+1$.
$\endgroup$ $\begingroup$The denominator has roots $i$ , $-1$ and $-i$ so you can write it in the following way $$ \frac{4x}{(x-i)(x+1)(x+i)}$$
Now separate it into partial fractions and integrate each of them individually .
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