Problem 18
Question
Use partial fractions to find the integral. $$ \int \frac{x^{2}-4 x+7}{x^{3}-x^{2}+x+3} d x $$
Step-by-Step Solution
Verified Answer
The integral is calculated by initially demonstrating the given fraction in the partial fractions form with A, B, and C as constants, where their values are calculated by factorizing the denominator and setting x to suitable values leading to equations in terms of A, B, and C. After determining these constants, the integral is computed. Considering the integral for \(\frac{1}{x+a}\) and the method for the integration of a quadratic polynomial could be helpful in computing the integral.
1Step 1: Decompose the fraction into partial fractions
The denominator of the fraction is a cubic polynomial, \( x^{3} - x^{2} + x + 3 \), and it can be factored into \((x+1)(x^2-2x+3)\). So, the fraction can be expressed as: \(\frac{x^{2}-4 x+7}{(x+1)(x^2-2x+3)} = \frac{A}{x+1} + \frac{Bx+C}{x^2-2x+3}\). Your task is now to find constants A,B and C.
2Step 2: Compute the constants A, B and C
To compute A,B, and C, multiply both sides by the common denominator \((x+1)(x^2-2x+3)\) and then set x to suitable values to get linear equations in terms of A, B and C. Solve these equations to find the values of A, B and C.
3Step 3: Integrate term by term
Once you have expressed the given fraction in terms of partial fractions, you can integrate term by term. Note that the integral of a sum is the sum of integrals.
4Step 4: Compute the integral
Compute the integral of each fraction separately. Remember, the integral of \(\frac{1}{x+a}\) is \(ln|x+a|\), and for the second fraction, use the formula for the integral of a quadratic polynomial.
Key Concepts
Partial FractionsIntegration TechniquesCubic Polynomials
Partial Fractions
Partial fractions are a method used in integral calculus to break down complex rational expressions into simpler ones. This technique is particularly useful when trying to integrate expressions with a polynomial in the denominator. By expressing complex fractions as a sum of simpler fractions, each of these fractions can be tackled individually for integration.
When dealing with a fraction that has a cubic polynomial in the denominator, like in our original exercise, we start by factoring the denominator. This could involve breaking it down into linear and irreducible quadratic factors.
This decomposition simplifies the integration process by allowing each term to be integrated separately, yielding the overall integral of the original fraction.
When dealing with a fraction that has a cubic polynomial in the denominator, like in our original exercise, we start by factoring the denominator. This could involve breaking it down into linear and irreducible quadratic factors.
- For instance, the denominator \(x^3 - x^2 + x + 3\) can be factored into \((x+1)(x^2-2x+3)\).
- The next step is to express the fraction as a sum of simpler fractions: \(\frac{A}{x+1} + \frac{Bx+C}{x^2-2x+3}\).
This decomposition simplifies the integration process by allowing each term to be integrated separately, yielding the overall integral of the original fraction.
Integration Techniques
Integration techniques are essential tools in calculus for finding antiderivatives. They encompass a wide range of methods, each applicable to different types of functions. When faced with a rational function such as the one involving partial fractions, certain techniques become particularly important.
One primary technique is the integration of basic rational functions. For instance:
By applying these techniques to the partial fraction decomposition, one can integrate each term individually, combining the results to find the integral of the original complex fraction. This step-by-step method not only makes calculations easier but also helps in gaining a clearer understanding of the function's behavior across its domain.
One primary technique is the integration of basic rational functions. For instance:
- The integral of \(\frac{1}{x+a}\) is \(\ln|x+a| + C\), where \(C\) is the constant of integration.
By applying these techniques to the partial fraction decomposition, one can integrate each term individually, combining the results to find the integral of the original complex fraction. This step-by-step method not only makes calculations easier but also helps in gaining a clearer understanding of the function's behavior across its domain.
Cubic Polynomials
Cubic polynomials are polynomials of degree three and can be expressed in the form \(ax^3 + bx^2 + cx + d\). These polynomials have distinct characteristics and can be more challenging to work with, especially when factoring. Factoring a cubic polynomial is the first step in partial fraction decomposition, as seen in the original exercise.
In some cases, a cubic polynomial can be factored into a product of linear factors and irreducible quadratic factors. For example:
Understanding how to manipulate cubic polynomials, either through factoring by grouping or using the Remainder Theorem, is pivotal. These methods will often lead to finding one or more of the roots, simplifying the polynomial further, and making the task of integration using partial fractions much more approachable.
In some cases, a cubic polynomial can be factored into a product of linear factors and irreducible quadratic factors. For example:
- The expression \(x^3 - x^2 + x + 3\) factors down to \((x+1)(x^2-2x+3)\).
Understanding how to manipulate cubic polynomials, either through factoring by grouping or using the Remainder Theorem, is pivotal. These methods will often lead to finding one or more of the roots, simplifying the polynomial further, and making the task of integration using partial fractions much more approachable.
Other exercises in this chapter
Problem 18
In Exercises \(7-26,\) evaluate the limit, using \(L\) 'Hôpital's Rule if necessary. (In Exercise \(12, n\) is a positive integer.) \(\lim _{x \rightarrow \inft
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Find the integral. (Note: Solve by the simplest method-not all require integration by parts.) $$ \int \frac{x}{\sqrt{2+3 x}} d x $$
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Find the integral involving secant and tangent. $$ \int \sec ^{4} 5 x d x $$
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