Evaluate Integral 4X/(X^3+X^2+X+1), Evaluate The Integral

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 $A=-2$.

Đang xem: Integral 4x/(x^3+x^2+x+1)

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 $u=x^2+1$.

answered Dec 4 “13 at 4:54


André NicolasAndré Nicolas
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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 .

answered Dec 4 “13 at 4:41


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