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Calculate. \int\frac{5 - x}{2x^2 + x - 1}

Question:

Calculate.

{eq}\int\frac{5 - x}{2x^2 + x - 1} {/eq}

Partial Fraction Integration:

A strategy to break down large and complex rational expressions into simpler rational expressions that add up to the equivalent original expression is partial fraction decomposition. In the example shown that when a binomial in the denominator is raised to a power such as {eq}\left(x+4\right)^2 {/eq}, the terms in the partial fraction decomposition change to {eq}\frac{A}{\left(x+4\right)} + \frac{B}{\left(x+4\right)^2} {/eq}, where {eq}A {/eq} and {eq}B {/eq} are based on the degree of the denominator, which should be algebraically adjusted by subtracting one degree so that the partial fraction decomposition is equivalent to the complete rational expression. In this case, {eq}A {/eq} and {eq}B {/eq} are constants.

Answer and Explanation:

Given: {eq}\int \frac{5-x}{2x^2+x-1} dx {/eq}


The first step is to find the partial fraction decomposition of {eq}\frac{5-x}{2x^2+x-1} {/eq}.


{eq}\begin{align*} \frac{5-x}{2x^2+x-1} &\Rightarrow \frac{5-x}{\left(2x-1\right)\left(x+1\right)} = \frac{A}{2x-1}+\frac{B}{x+1} \\ &\Rightarrow \frac{\left(5-x\right)\left(2x-1\right)\left(x+1\right)}{\left(2x-1\right)\left(x+1\right)} = \frac{A\left(2x-1\right)\left(x+1\right)}{2x-1}+\frac{B\left(2x-1\right)\left(x+1\right)}{x+1} \\ &\Rightarrow 5-x = A\left(x+1\right)+B\left(2x-1\right) \\ &\Rightarrow 5-(-1) = A\left((-1)+1\right)+B\left(2(-1)-1\right) \text{ [When x = -1]} \\ &\Rightarrow 5+1 = A\left(0\right)+B\left(-2-1\right) \\ &\Rightarrow 6 = 0+B\left(-3\right) \\ &\Rightarrow 6 = -3B \\ &\Rightarrow \frac{-1}{3}\cdot 6 = \frac{-1}{3}\cdot -3B \\ &\Rightarrow B = -2 \\ &\Rightarrow 5-(\frac{1}{2}) = A\left(\frac{1}{2}+1\right)+B\left(2(\frac{1}{2})-1\right) \text{ [When x = 1/2]} \\ &\Rightarrow \frac{10}{2}-\frac{1}{2} = A\left(\frac{1}{2}+\frac{2}{2}\right)+B\left(1-1\right) \\ &\Rightarrow \frac{9}{2} = A\left(\frac{3}{2}\right)+B\left(0\right) \\ &\Rightarrow \frac{9}{2} = \frac{3}{2}A+0 \\ &\Rightarrow \frac{9}{2} = \frac{3}{2}A \\ &\Rightarrow \frac{2}{3}\cdot \frac{9}{2} = \frac{2}{3}\cdot \frac{3}{2}A \\ &\Rightarrow A = 3 \\ &\Rightarrow \frac{5-x}{\left(2x-1\right)\left(x+1\right)} = \frac{3}{2x-1}+\frac{-2}{x+1} \end{align*} {/eq}.


Now this partial fraction can be used to evaluate the integral.


{eq}\begin{align*} \int \frac{5-x}{2x^2+x-1} dx &= \int \frac{3}{2x-1}-\frac{2}{x+1} dx \\ &= \int \frac{3}{2x-1}dx-\int \frac{2}{x+1}dx \text{ [Integral Sum Rule]} \\ &= 3\cdot \int \frac{1}{2x-1}dx-2\cdot \int \frac{1}{x+1}dx \\ &= 3\cdot \frac{1}{2}\ln \left|2x-1\right|-2\ln \left|x+1\right| \text{ [First Integral due to Integral Chain Rule]} \\ &= \frac{3}{2}\ln \left|2x-1\right|-2\ln \left|x+1\right|+c \text{ [Constant added due to indefinite integral (lack of bounds)]} \end{align*} {/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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