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Found in: Page 334

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# Write out the form of the partial fraction decomposition of the function (as in Example 6). Do not determine the numerical values of the coefficients.(a) $$\frac{{{x^4} - 2{x^3} + {x^2} + 2x - 1}}{{{x^2} - 2x + 1}}$$(b)$$\frac{{{x^2} - 1}}{{{x^3} + {x^2} + x}}$$

$\frac{{{x^2}}}{{{x^2} - 2x + 1}} + \frac{{2x}}{{{x^2} - 2x + 1}} - \frac{1}{{{x^2} - 2x + 1}} + \frac{{{x^4}}}{{{x^2} - 2x + 1}} - \frac{{2{u^3}}}{{{x^2} - 2x + 4}}$

See the step by step solution

## Step 1: Decomposition of the partial fraction.

$$\frac{{{x^4} - 2{x^3} + {x^2} + 2x - 1}}{{{x^2} - 2x + 1}}$$

$${x^2} - 2x + 1 = {\left( {x - 1} \right)^2}$$

## Step 2: Calculation of partial fraction.

$${x^4} - 2{x^3} + {x^2} + 2x - 1 = {x^2}\left( {{x^2} - 2x + 1} \right) + (2x - 1)$$

Written as

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

Expansion

$\frac{{{x^2}}}{{{x^2} - 2x + 1}} + \frac{{2x}}{{{x^2} - 2x + 1}} - \frac{1}{{{x^2} - 2x + 1}} + \frac{{{x^4}}}{{{x^2} - 2x + 1}} - \frac{{2{u^3}}}{{{x^2} - 2x + 4}}$

Hence, This is the final answer.

## (b) Step 1:

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

## (b) Step 2:  Calculation of Partial fraction.

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

Hence $$\frac{A}{x} + \frac{{Bx + c}}{{{x^2}(x + 1)}}$$ is final answer.