Find the partial fractions decomposition and an antiderivative. \frac{3x + 5}{5x^2 - 4x - 1}

Question:

Find the partial fractions decomposition and an antiderivative.

{eq}\frac{3x + 5}{5x^2 - 4x - 1} {/eq}

Integration Using Partial Fraction Decomposition:

Partial fraction decomposition of a fraction number can be inferred by glimpsing the factor in the denominator. Relying on the factor in the denominator, we compose the term in partial fraction decomposition.


When factor in the denominator is:

{eq}(ax+b) {/eq}

Then term in the partial fraction decomposition is:

{eq}\frac{A}{ax+b} {/eq}

When factor in the denominator is:

{eq}(ax+b)^k {/eq}

Then term in the partial fraction decomposition are:

{eq}\frac{A_1}{ax+b} + \frac{A_2}{(ax+b)^2} + \frac{A_3}{(ax+b)^3 } + ......+ \frac{A_k}{(ax+b)^k} \hspace{1 cm} \text{ k = 1,2,3.......} {/eq}

When factor in the denominator is:

{eq}(ax^2+bx+c) {/eq}

Then term in the partial fraction decomposition is:

{eq}\frac{Ax + B}{ax^2+bx+c} {/eq}

When factor in the denominator is:

{eq}(ax^2+bx+c)^k {/eq}

Then term in the partial fraction decomposition are:

{eq}\frac{A_1x + B_1}{ax^2+bx+c} + \frac{A_2x + B_2}{(ax^2+bx+c)^2} + \frac{A_3x + B_3}{(ax^2+bx+c)^3 } + ......+ \frac{A_kx + B_k}{(ax^2+bx+c)^k} \hspace{1 cm} \text{ k = 1,2,3.......} {/eq}

Answer and Explanation:

The fraction number is given by:

{eq}\frac{3x+5}{5x^2 - 4x - 1} {/eq}


Factorising the denominator in the given fraction number, we have:

{eq}\begin{align*} \frac{3x+5}{5x^2 - 4x - 1} &= \frac{3x+5}{5x^2 - 5x + x - 1} \\ &= \frac{3x+5}{5x(x-1) + 1(x-1)} \\ &= \frac{3x+5}{(x-1)(5x+1)} \end{align*} {/eq}


Terms in the partial fraction decomposition are:

{eq}\frac{3x+5}{(x-1)(5x+1)} = \frac{A}{x-1} + \frac{B}{5x + 1} \hspace{1 cm} \left[ \text{ Where A and B are constant numbers} \right] \\ \Rightarrow 3x + 5 = A(5x+1) + B(x-1) \hspace{1 cm} \text{(Equation 1)} {/eq}


For finding the constant terms {eq}A \, and \, B {/eq} , we have to substitute the root from each factor of the denominator.

Substituting the root of factor {eq}(x - 1) {/eq} which is {eq}1 {/eq} into the (Equation 1), we have:

{eq}3(1) + 5 = A\left( 5(1) + 1 \right) + B (1 - 1) \\ \Rightarrow 8 = 6A \\ \Rightarrow A = \frac{8}{6} \\ \Rightarrow A = \frac{4}{3} {/eq}


Again:

Substituting the root of factor {eq}(5x+1) {/eq} which is {eq}- \frac{1}{5} {/eq} into the (Equation 1), we have:

{eq}3(- \frac{1}{5}) + 5 = A\left( 5(- \frac{1}{5}) + 1 \right) + B (- \frac{1}{5} - 1) \\ \Rightarrow 5 - \frac{3}{5} = - \frac{6B}{5} \\ \Rightarrow \frac{22}{5} = - \frac{6B}{5} \\ \Rightarrow B = - \frac{22}{6} \\ \Rightarrow B = - \frac{11}{3} {/eq}


So:

The partial fraction decomposition of the given fraction is:

{eq}\frac{3x+5}{5x^2 - 4x - 1} = \frac{4}{3(x-1)} - \frac{11}{3(5x + 1)} \hspace{1 cm} \left[ \because A = \frac{4}{3} \, and \, B = - \frac{11}{3} \right] {/eq}


Now:

Taking the integration into the partial fraction decomposition, we have:

{eq}\int \frac{3x+5}{5x^2 - 4x -1} \, dx = \frac{4}{3} \int \frac{dx}{x-1} - \frac{11}{3} \int \frac{dx}{5x+1} {/eq}

Let us assume that:

{eq}x - 1 = u \\ \Rightarrow dx = du {/eq}

And:

{eq}5x + 1 = v \\ \Rightarrow 5 \, dx = dv \\ \Rightarrow dx = \frac{dv}{5} {/eq}


Substituting {eq}u \, and \, v {/eq} into the integration and we have:

{eq}\begin{align*} \int \frac{3x+5}{5x^2 - 4x -1} \, dx &= \frac{4}{3} \int \frac{du}{u} - \frac{11}{15} \int \frac{dv}{v} \\ &= \frac{4}{3} \ln \mid u \mid - \frac{11}{15} \ln \mid v \mid + C \hspace{1 cm} \left[ \because \int \frac{dx}{x} = \ln \mid x \mid + C \, \text{( Where C is a constant of integration)} \right] \end{align*} {/eq}


Undo substitution and we have:

{eq}\int \frac{3x+5}{5x^2 - 4x - 1} \, dx = \frac{4}{3} \ln \mid x-1 \mid - \frac{11}{15} \ln \mid 5x+1 \mid + C \hspace{1 cm} \left[ \because u = x - 1 \, and \, v = 5x + 1 \right] {/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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