Find the partial fractions decomposition and an antiderivative. \frac{4x - 5}{x^3 - 3x^2}

Question:

Find the partial fractions decomposition and an antiderivative.

{eq}\frac{4x - 5}{x^3 - 3x^2} {/eq}

Partial Fractions:

Decomposition into partial fractions is necessary for the decomposition of complex rational expressions.

With them, operators such as integrals are easier to operate.

Answer and Explanation:

First, we obtain the irreductible factors for the denominator:

{eq}\frac{{4x - 5}}{{{x^3} - 3{x^2}}} = \frac{{4x - 5}}{{{x^2}\left( {x - 3} \right)}} {/eq}

In partial fractions the rational expression can be written as:

{eq}\frac{{4x - 5}}{{{x^2}\left( {x - 3} \right)}} = \frac{A}{x} + \frac{B}{{{x^2}}} + \frac{C}{{x - 3}}\\ \frac{{4x - 5}}{{{x^2}\left( {x - 3} \right)}} = \frac{{Ax\left( {x - 3} \right) + B\left( {x - 3} \right) + C{x^2}}}{{{x^2}\left( {x - 3} \right)}} {/eq}

Equating numerators, we calculate the constants:

{eq}4x - 5 = Ax\left( {x - 3} \right) + B\left( {x - 3} \right) + C{x^2}\\ 4x - 5 = A{x^2} - 3Ax + Bx - 3B + C{x^2}\\ \\ \left\{ \begin{array}{l} 0 = A + C\\ 4 = - 3A + B\\ - 5 = - 3B \end{array} \right.\\ \\ A = - 7/9,B = 5/3,C = 7/9 {/eq}

So, the decomposition can be written as: {eq}\frac{{4x - 5}}{{{x^2}\left( {x - 3} \right)}} = \frac{{ - 7/9}}{x} + \frac{{5/3}}{{{x^2}}} + \frac{{7/9}}{{x - 3}}. {/eq}


Learn more about this topic:

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How to Integrate Functions With Partial Fractions

from Math 104: Calculus

Chapter 13 / Lesson 9
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