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

## Question:

Find the partial fractions decomposition and an antiderivative.

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

## Partial Fraction Concept:

Let{eq}A(x) {/eq} can be written as {eq}\frac{{d(x)}}{{g(x)}} {/eq} ,in which both are polynomials in x.In partial fraction degree of {eq}g(x) {/eq} must be greater than {eq}d(x) {/eq}then partial fraction for linear and quadratic the factor given by;

{eq}A(x) = \frac{{d(x)}}{{(x + p)({x^2} + ax + b)}} {/eq}

By partial fraction expansion ,

{eq}A(x) = \frac{W}{{(x + p)}} + \frac{{Bx + C}}{{({x^2} + ax + b)}} {/eq}

The following rules are relevant to this problem:

1.{eq}{\int {\left( {e\left( x \right) + y\left( x \right)} \right)dx = \int {e\left( x \right)dx} + \int {y\left( x \right)dx} } } {/eq}

2.{eq}{\int {cy\left( x \right)dx = c\int {y\left( x \right)dx} } } {/eq}

3.{eq}{\int {\frac{1}{x}dx = \ln \left| x \right| + c} } {/eq}

## Answer and Explanation:

Given that: {eq}\displaystyle \frac{{5x - 2}}{{{x^2} - 4}} {/eq}

{eq}\displaystyle \eqalign{ & \frac{{5x - 2}}{{{x^2} - 4}} \cr & \frac{{5x - 2}}{{\left( {x + 2} \right)\left( {x - 2} \right)}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{x^2} - 4 = \left( {x + 2} \right)\left( {x - 2} \right)} \right) \cr & {\text{Let,}} \cr & f\left( x \right) = \frac{{5x - 2}}{{{x^2} - 4}} \cr & \frac{{5x - 2}}{{{x^2} - 4}} = \frac{A}{{x + 2}} + \frac{B}{{x - 2}} \cr & 5x - 2 = A\left( {x - 2} \right) + B\left( {x + 2} \right) \cr & {\text{Putting }}x = 2; \cr & 10 - 2 = 0 + B\left( 4 \right) \cr & 8 = B\left( 4 \right) \cr & B = 2 \cr & \cr & {\text{Putting }}x = - 2; \cr & - 10 - 2 = A\left( { - 4} \right) + 0 \cr & - 12 = A\left( { - 4} \right) \cr & A = 3 \cr & \cr & f\left( x \right) = \frac{{5x - 2}}{{{x^2} - 4}} = \frac{3}{{x + 2}} + \frac{2}{{x - 2}} \cr & \cr & \int {\frac{{5x - 2}}{{{x^2} - 4}}dx} \cr & \int {\left( {\frac{3}{{x + 2}} + \frac{2}{{x - 2}}} \right)dx} \cr & \int {\left( {\frac{3}{{x + 2}}} \right)dx} + \int {\left( {\frac{2}{{x - 2}}} \right)dx} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\int {\left( {e\left( x \right) + y\left( x \right)} \right)dx = \int {e\left( x \right)dx} + \int {y\left( x \right)dx} } } \right) \cr & 3\int {\frac{1}{{x + 2}}dx} + 2\int {\frac{1}{{x - 2}}dx} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\int {cy\left( x \right)dx = c\int {y\left( x \right)dx} } } \right) \cr & 3\ln \left| {x + 2} \right| + 2\ln \left| {x - 2} \right| + c\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\int {\frac{1}{x}dx = \ln \left| x \right| + c} } \right) \cr & \cr & \int {\frac{{5x - 2}}{{{x^2} - 4}}dx} = 3\ln \left| {x + 2} \right| + 2\ln \left| {x - 2} \right| + c \cr} {/eq}

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from High School Algebra I: Help and Review

Chapter 3 / Lesson 25