Problem 36
Question
Use partial-fraction decomposition to evaluate the integrals. $$ \int \frac{4 x^{2}-x-1}{(x+1)(x-3)} d x $$
Step-by-Step Solution
Verified Answer
The integral evaluates to \(\frac{5}{4} \ln |x+1| + \frac{11}{4} \ln |x-3| + C\).
1Step 1: Decompose the Rational Function
Start by assuming a form for the partial fraction decomposition. The denominator \((x+1)(x-3)\) suggests that our decomposition will be of the form:\[\frac{A}{x+1} + \frac{B}{x-3}\]So, we need the equation:\[\frac{4x^2 - x - 1}{(x+1)(x-3)} = \frac{A}{x+1} + \frac{B}{x-3}\]Multiply through by the denominator \((x+1)(x-3)\) to clear it from the fractions:\[4x^2 - x - 1 = A(x-3) + B(x+1)\].
2Step 2: Expand and Collect Like Terms
Expand the right-hand side of the equation:\[A(x-3) = Ax - 3A\]\[B(x+1) = Bx + B\]Combine them:\[Ax - 3A + Bx + B = (A + B)x + (-3A + B)\].
3Step 3: Equate Coefficients to Solve for A and B
Match the coefficients from both sides of the equation. The original left side is:\[4x^2 - x - 1\]The expanded right-hand side provides:\[(A + B)x + (-3A + B)\]Equate coefficients of similar terms:1. From \(x\):\[A + B = 4\]2. From constant terms (\(x^0\)):\[-3A + B = -1\]
4Step 4: Solve the System of Equations
Solve the equations:1. \(A + B = 4\)2. \(-3A + B = -1\)Substitute the first into the second:\[B = 4 - A\]\[-3A + (4 - A) = -1\]This simplifies to:\[-4A + 4 = -1\]Solving for A:\[-4A = -5\]\[A = \frac{5}{4}\]Solve for B:\[B = 4 - A = 4 - \frac{5}{4} = \frac{16}{4} - \frac{5}{4} = \frac{11}{4}\].
5Step 5: Integrate Each Partial Fraction
Now substitute back to integrate:\[\int \frac{5}{4(x+1)}\, dx + \int \frac{11}{4(x-3)}\, dx\]This becomes:\[\frac{5}{4}\int \frac{1}{x+1} \, dx + \frac{11}{4}\int \frac{1}{x-3} \, dx\]The integrals are:\[\frac{5}{4} \ln |x+1| + \frac{11}{4} \ln |x-3| + C\]
Key Concepts
Integration TechniquesRational FunctionsSystem of Equations
Integration Techniques
Integration is a key technique in calculus, used to find the area under curves or solve differential equations. One sophisticated method of integration is partial-fraction decomposition. This technique is especially useful for integrating rational functions.
It involves breaking down a complex fraction into simpler, more manageable parts. By expressing the original fraction as a sum of simpler fractions, each piece is easier to integrate individually.
In our exercise, the integral of a complex fraction \[ \int \frac{4x^2 - x -1}{(x+1)(x-3)} \, dx \]is simplified using partial-fraction decomposition into separate terms. This transforms the problem into simpler integrals to evaluate, like:
It involves breaking down a complex fraction into simpler, more manageable parts. By expressing the original fraction as a sum of simpler fractions, each piece is easier to integrate individually.
In our exercise, the integral of a complex fraction \[ \int \frac{4x^2 - x -1}{(x+1)(x-3)} \, dx \]is simplified using partial-fraction decomposition into separate terms. This transforms the problem into simpler integrals to evaluate, like:
- \( \int \frac{5}{4(x+1)} \, dx \)
- \( \int \frac{11}{4(x-3)} \, dx \)
Rational Functions
Rational functions are fractions in which both the numerator and the denominator are polynomials. They can be written in the general form:\[ \frac{P(x)}{Q(x)} \]In our example, \[ \frac{4x^2 - x - 1}{(x+1)(x-3)} \] is a rational function with a quadratic polynomial in the numerator and a product of linear polynomials in the denominator.
To work with rational functions via integration or other calculus operations, it can be helpful to break the fraction down using partial-fraction decomposition.
This process requires expressing the original fraction as a sum of fractions, whose denominators are simpler polynomials usually taken from the factors of the original denominator.
To work with rational functions via integration or other calculus operations, it can be helpful to break the fraction down using partial-fraction decomposition.
This process requires expressing the original fraction as a sum of fractions, whose denominators are simpler polynomials usually taken from the factors of the original denominator.
System of Equations
Systems of equations come into play when finding unknown constants in partial-fraction decomposition. In this method, we set equations based on matching coefficients from expanded polynomials.
A system of equations is a collection of two or more equations with common variables. In our problem, after assigning a form for partial fractions, \[ \frac{A}{x+1} + \frac{B}{x-3} \] we equated coefficients to derive the following system:
This step is crucial as it determines the specific values needed to decompose the original integral, allowing for its evaluation.
A system of equations is a collection of two or more equations with common variables. In our problem, after assigning a form for partial fractions, \[ \frac{A}{x+1} + \frac{B}{x-3} \] we equated coefficients to derive the following system:
- \( A + B = 4 \)
- \( -3A + B = -1 \)
This step is crucial as it determines the specific values needed to decompose the original integral, allowing for its evaluation.
Other exercises in this chapter
Problem 36
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