Consider the indefinite integral. \int\frac{6x^3+3x^2-19x-10}{x^2-4}\ dx Then the integrand...

Question:

Consider the indefinite integral.

{eq}\int\frac{6x^3+3x^2-19x-10}{x^2-4}\ dx {/eq}

Then the integrand decomposes in the form {eq}ax + b + \frac{c}{x-2} + \frac{d}{x+2} {/eq}

Compute the coefficients and evaluate the integral.

Integrals Using Partial Fraction Decomposition:

As much as we want simpler expressions, integration tends to favor expanded expressions. For example, if we want to integrate a factored polynomial, the approach we take is to multiply out the polynomial and take the integral of each term. In this solution, we are given a rational function, and since integrals are easier with expanded expressions, we apply long division and express the remainder using partial fraction decomposition.

Answer and Explanation:

We are asked to evaluate the integral $$\int \frac{6x^3 + 3x^2 - 19x - 10}{x^2 - 4} \, \mathrm{d}x $$ Note that the integrand is a rational function where the degree of the numerator is higher than that of the denominator, which means division is possible to simplify the integrand. By long division, we have that $$\frac{6x^3 + 3x^2 - 19x - 10}{x^2 - 4} = 6x + 3 + \frac{5x + 2}{x^2 - 4} $$ which gives us an expression similar to the one in the question. However, the fraction on the right side can still be decomposed into partial fractions. We have that {eq}x^2 - 4 = (x - 2)(x + 2) {/eq} so a partial decomposition of {eq}\dfrac{5x + 2}{x^2 - 4} {/eq} is of the form $$\frac{5x + 2}{x^2 - 4} = \frac{c}{x - 2} + \frac{d}{x + 2} $$

We then have $$\begin{align*} \frac{5x + 2}{x^2 - 4} &= \frac{c}{x - 2} + \frac{d}{x + 2} \\ 5x + 2 &= c(x + 2) + d(x - 2) \end{align*} $$

Taking {eq}x = 2 {/eq}, we have $$\begin{align*} 12 &= 4c \\ c &= 3 \end{align*} $$

Taking {eq}x = -2 {/eq}, we have $$\begin{align*} -8 &= -4d \\ d &= 2 \end{align*} $$ and so the integrand can then be rewritten as $$\frac{6x^3 + 3x^2 - 19x - 10}{x^2 - 4} = 6x + 3 + \frac{3}{x - 2} + \frac{2}{x + 2} $$ so the coefficients are given by {eq}\boxed{(a, b, c, d) = (6, 3, 3, 2)} {/eq}.

We can then compute the integral using this new form as $$\begin{align*} \int \frac{6x^3 + 3x^2 - 19x - 10}{x^2 - 4} \, \mathrm{d}x &= \int \left(6x + 3 + \frac{3}{x - 2} + \frac{2}{x + 2} \right) \, \mathrm{d}x \\ &= \boxed{3x^2 + 3x + 3\ln|x - 2| + 2\ln|x + 2| + C} \end{align*} $$


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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