# 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\$.

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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 .

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answered Dec 4 “13 at 4:41

abkdsabkds
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