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Chapter 6: Problem 20

Use partial fractions to find the indefinite integral. $$ \int \frac{5}{x^{2}+x-6} d x $$

Short Answer

Expert verified
After all, the indefinite integral \( \int \frac{5}{x^{2}+x-6} dx \) can be expressed as \( A \times ln |x - 3| + B \times ln |x + 2| + C \) after making use of partial fraction decomposition. Note that specific values of A, B and C depend on the solution of the equation in step 2.

Step by step solution

01

Factorize the Denominator

In order to express the function in the form of partial fractions, the denominator \(x^2 + x - 6\) must be factorized. A polynomial of this type can often be expressed as the product of two binomial terms: \( (x - a)(x - b) \). To find the values of \(a\) and \(b\), look for two numbers that multiply to -6 (the constant term) and add to 1 (the coefficient of \(x\)). The suitable values here are 3 and -2, they multiply to -6 and add up to 1. So, the denominator can be factorized as \( (x - 3)(x + 2) \).
02

Express as Partial Fractions

Next, write the original function as a sum of partial fractions: \( \frac{A}{x-3} + \frac{B}{x+2} = \frac{5}{x^2+x-6}\) . Both sides of the equation must equate for all values of \(x\). Therefore, by getting rid of the denominator, the equation can be simplifies to: \( A(x+2) + B(x - 3) = 5 \). Now, find the constants \(A\) and \(B\) by using suitable values of \(x\). For instance, one might set \(x = 3\) or \(x = -2\), because these would make either \(A\) or \(B\) term in the equation vanish. Then, solve the equations for corresponding \(A\) and \(B\).
03

Integration of Partial Fractions

With the function now expressed in terms of partial fractions, each fraction can be integrated separately. For each part, apply the rule of integral for 1/x, which is \( ln |x| \). So, the integral for each term is \( A \times \int \frac{1}{x-3} dx + B \times \int \frac{1}{x+2} dx \). Evaluate these integrals. The integral of the first fraction will result in \( A \times ln |x - 3| \) and the integral of the second fraction will result in \( B \times ln |x + 2| \).
04

Combine and Finalize Solution

Combine the integral results of the separate fractions, this gives: \( \int \frac{5}{x^{2}+x-6} dx = A \times ln |x - 3| + B \times ln |x + 2| + C \). The \(C\) is the constant of integration that appears when we integrate.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Indefinite Integral
An indefinite integral is essentially the reverse process of differentiation. When we perform integration, we are finding a function whose derivative is the given function. For example, integrating a function helps us find the area under a curve for any given interval. In this context, it's useful to remember a few key ideas:
  • Unlike a definite integral, an indefinite integral does not have specific limits; thus, the result will always include a constant of integration, often noted as 'C'.
  • The integral sign ( \( \int \)) serves as a command to find the antiderivative of the given function.
  • Expressions like 'dy/dx' hint at what our goal is: undoing the effect of differentiation on some starting function to retrieve a more general form.
It's crucial to understand that each indefinite integral result can represent potentially infinite forms due to this constant 'C', reflecting a family of curves.
Polynomial Factorization
To successfully integrate some rational expressions, we use polynomial factorization. This means rewriting a complex polynomial into products of smaller and simpler polynomials. In our exercise, the denominator of the expression makes it perfect for us to use factorization:
  • Here, the polynomial \( x^2 + x - 6 \) is factorized into \( (x - 3)(x + 2) \). This step sets the stage for using partial fractions.
  • The choice of factor pairs stems from finding numerical values that multiply to give the constant term (-6) and add up to the middle coefficient (+1).
Once factorized, these components allow the expression to disassemble into simpler fractions where the process of integration becomes simplified.
Natural Logarithm
Natural logarithms are pivotal when integrating functions that yield fractions of the form \( \frac{1}{x-a} \). Here's how they come into play with our integral solution:
  • Integrating a term like \( \frac{1}{x-3} \) results in \( \ln |x-3| \), and integrating \( \frac{1}{x+2} \) gives us \( \ln |x+2| \).
  • Natural logs help us derive integrals where exponents or polynomials feature prominently by transforming the expression into one that's easier to handle algebraically.
It's also good to understand that, in this context, the absolute value around the 'x' terms ensures that the logarithmic function remains defined for all real numbers within the integration spectrum.

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Most popular questions from this chapter

Approximate the integral using (a) the Trapezoidal Rule and (b) Simpson's Rule for the indicated value of \(n .\) (Round your answers to three significant digits.) $$ \int_{0}^{1} \sqrt{1-x^{2}} d x, n=8 $$

Use a spreadsheet to complete the table for the specified values of and to demonstrate that $$ \lim _{x \rightarrow \infty} x^{n} e^{-a x}=0, \quad a>0, n>0 $$ $$ \begin{array}{|c|c|c|c|c|}\hline x & {1} & {10} & {25} & {50} \\ \hline x^{n} e^{-a x} & {} & {} & {} & {} \\ \hline\end{array} $$ $$ a=1, n=1 $$

Use the error formulas to find bounds for the error in approximating the integral using (a) the Trapezoidal Rule and (b) Simpson's Rule. (Let \(n=4 .)\) $$ \int_{0}^{2} x^{3} d x $$

Use a spreadsheet to complete the table for the specified values of and to demonstrate that $$ \lim _{x \rightarrow \infty} x^{n} e^{-a x}=0, \quad a>0, n>0 $$ $$ \begin{array}{|c|c|c|c|c|}\hline x & {1} & {10} & {25} & {50} \\ \hline x^{n} e^{-a x} & {} & {} & {} & {} \\ \hline\end{array} $$ $$ a=\frac{1}{2}, n=2 $$

Use a spreadsheet to complete the table for the specified values of and to demonstrate that $$ \lim _{x \rightarrow \infty} x^{n} e^{-a x}=0, \quad a>0, n>0 $$ $$ \begin{array}{|c|c|c|c|c|}\hline x & {1} & {10} & {25} & {50} \\ \hline x^{n} e^{-a x} & {} & {} & {} & {} \\ \hline\end{array} $$ $$ a=2, n=4 $$
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