How do I find the partial fraction decomposition of $\frac{1}{x^{3} + 2 x^{2} + x}$ $(1)/(x^3+2x^2+x$ ?

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How do I find the partial fraction decomposition of $\frac{1}{x^{3} + 2 x^{2} + x}$ $(1)/(x^3+2x^2+x$ ?

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Answer 1 · 2018-03-14T23:54:46

$\frac{1}{x} - \frac{1}{x + 1} - \frac{1}{{(x + 1)}^{2}}$ $1/x-1/(x+1)-1/(x+1)^2$

Explanation:

$\frac{1}{x^{3} + 2 x^{2} + x} = \frac{1}{x \cdot (x^{2} + 2 x + 1)} = \frac{1}{x \cdot {(x + 1)}^{2}}$ $1/(x^3+2x^2+x)=1/[x*(x^2+2x+1)]=1/[x*(x+1)^2]$

Hence,

$\frac{1}{x^{3} + 2 x^{2} + x} = \frac{A}{x} + \frac{B}{x + 1} + \frac{C}{{(x + 1)}^{2}}$ $1/(x^3+2x^2+x)=A/x+B/(x+1)+C/(x+1)^2$

After expanding denominator,

$A \cdot {(x + 1)}^{2} + B \cdot (x^{2} + x) + C x = 1$ $A*(x+1)^2+B*(x^2+x)+Cx=1$

Set $x = - 1$ $x=-1$ , $- C = 1$ $-C=1$ , so $C = - 1$ $C=-1$

Set $x = 0$ $x=0$ , $A = 1$ $A=1$

Set $x = 1$ $x=1$ , $- 4 A + 2 B + C = 1$ $-4A+2B+C=1$ , so $B = - 1$ $B=-1$

Thus,

$\frac{1}{x^{3} + 2 x^{2} + x} = \frac{1}{x} - \frac{1}{x + 1} - \frac{1}{{(x + 1)}^{2}}$ $1/(x^3+2x^2+x)=1/x-1/(x+1)-1/(x+1)^2$

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How do I find the partial fraction decomposition of $\frac{1}{x^{3} + 2 x^{2} + x}$ $(1)/(x^3+2x^2+x$ ?

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Explanation:

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