# 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.

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}