# How do I find the partial fraction decomposition of (1)/(x^3+2x^2+x ?

Mar 14, 2018

$\frac{1}{x} - \frac{1}{x + 1} - \frac{1}{x + 1} ^ 2$

#### Explanation:

$\frac{1}{{x}^{3} + 2 {x}^{2} + x} = \frac{1}{x \cdot \left({x}^{2} + 2 x + 1\right)} = \frac{1}{x \cdot {\left(x + 1\right)}^{2}}$

Hence,

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

After expanding denominator,

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

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

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

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

Thus,

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