Chapter 8: Problem 29
In Exercises \(21-32,\) express the integrand as a sum of partial fractions and evaluate the integrals. $$\int \frac{x^{2}}{x^{4}-1} d x$$
Short Answer
Expert verified
The integral evaluates to \(0.5 \ln|x^2 - 1| + C\).
Step by step solution
01
Factor the Denominator
First, we need to factor the denominator \(x^4 - 1\). We can recognize \(x^4 - 1\) as a difference of squares: \((x^2)^2 - 1^2\), which factors into \((x^2 - 1)(x^2 + 1)\). Furthermore, \(x^2 - 1\) can be factored as \((x - 1)(x + 1)\). Thus, the complete factorization is \((x - 1)(x + 1)(x^2 + 1)\).
02
Setup Partial Fractions
Express the integrand \(\frac{x^2}{(x - 1)(x + 1)(x^2 + 1)}\) as a sum of partial fractions: \[ \frac{A}{x - 1} + \frac{B}{x + 1} + \frac{Cx + D}{x^2 + 1} \] where \(A\), \(B\), \(C\), and \(D\) are constants to be determined.
03
Clear the Denominator
Multiply both sides by the common denominator \((x - 1)(x + 1)(x^2 + 1)\) to eliminate the fractions: \[ x^2 = A(x + 1)(x^2 + 1) + B(x - 1)(x^2 + 1) + (Cx + D)(x^2 - 1) \].
04
Expand and Equate Coefficients
Expand the right-hand side: \[ A(x^3 + x + x^2 + 1) + B(x^3 - x + x^2 - 1) + Cx(x^2 - 1) + D(x^2 - 1) \]. Now group and equate coefficients of like powers of \(x\) from both sides to find the values of \(A\), \(B\), \(C\), and \(D\).
05
Solve for Constants
Set up a system of equations by equating coefficients of \(x^3\), \(x^2\), \(x\), and the constant terms: \[ A + B + C = 0 \] \[ A + B = 1 \] \[ D = 0 \] \[ A - B - D = 0 \]. Solving the system, we find \(A = 0.5\), \(B = 0.5\), \(C = 0\), and \(D = 0\).
06
Integrate Each Term
Now integrate each term separately: \[ \int \frac{0.5}{x - 1} dx + \int \frac{0.5}{x + 1} dx \]. The terms involving \(Cx + D\) vanish because \(C = 0\) and \(D = 0\).
07
Solve the Integrals
Calculate the integrals: \( \int \frac{0.5}{x - 1} dx = 0.5 \ln|x - 1| + C_1 \) and \( \int \frac{0.5}{x + 1} dx = 0.5 \ln|x + 1| + C_2 \). Combine the constants for simplicity as \(C\).
08
Combine Results
The integral of the original expression is: \[ 0.5 (\ln|x - 1| + \ln|x + 1|) + C \]. This can be further simplified using logarithm properties: \(0.5 \ln|x^2 - 1| + C\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration Techniques
When dealing with integrals that involve rational expressions, like \( \int \frac{x^{2}}{x^{4}-1} dx \), it's useful to break down the process using specific techniques. One key technique is partial fraction decomposition. This involves expressing a rational function as a sum of simpler fractions, which are easier to integrate.
For the integral given, the integrand is \( \frac{x^2}{x^4 - 1} \). To integrate this function, we first decompose it into partial fractions. This makes the integration of each separate term a straightforward process.
For the integral given, the integrand is \( \frac{x^2}{x^4 - 1} \). To integrate this function, we first decompose it into partial fractions. This makes the integration of each separate term a straightforward process.
- We begin by factoring the denominator, recognizing it as a difference of squares.
- After factoring, each simpler fraction can be integrated separately.
- This technique is particularly useful for complex denominators appearing in a rational function.
Polynomial Factorization
A crucial step in partial fraction decomposition involves factoring the polynomial in the denominator. For our given integral, the denominator is \( x^4 - 1 \). This expression is factored using the identity for the difference of squares.
We initially rewrite \( x^4 - 1 \) as \((x^2)^2 - 1^2 \), which is directly factored into \((x^2 - 1)(x^2 + 1) \). Further simplification of \( x^2 - 1 \) gives us \((x - 1)(x + 1) \). Thus, the full factorization is \((x - 1)(x + 1)(x^2 + 1) \).
We initially rewrite \( x^4 - 1 \) as \((x^2)^2 - 1^2 \), which is directly factored into \((x^2 - 1)(x^2 + 1) \). Further simplification of \( x^2 - 1 \) gives us \((x - 1)(x + 1) \). Thus, the full factorization is \((x - 1)(x + 1)(x^2 + 1) \).
- Factoring informs us about possible partial fraction forms.
- It reduces the complexity of the rational expression.
- This step is imperative for isolating simpler fractions.
Logarithmic Integration
Once we have decomposed a rational function into partial fractions, integrating terms often leads to logarithmic forms. For the integral \( \int \frac{x^{2}}{x^{4}-1} dx \), and its partial fractions, logarithmic integration is a straightforward application.
When you encounter fractions of the form \( \frac{A}{x - a} \), its integral is \( A \ln|x - a| + C \). In our exercise, terms such as \( \int \frac{0.5}{x-1} dx \) and \( \int \frac{0.5}{x+1} dx \) simplify to logarithmic functions:
When you encounter fractions of the form \( \frac{A}{x - a} \), its integral is \( A \ln|x - a| + C \). In our exercise, terms such as \( \int \frac{0.5}{x-1} dx \) and \( \int \frac{0.5}{x+1} dx \) simplify to logarithmic functions:
- \( \int \frac{0.5}{x-1} dx = 0.5 \ln|x-1| + C_1 \)
- \( \int \frac{0.5}{x+1} dx = 0.5 \ln|x+1| + C_2 \)