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Magazine Chapter 7: Problem 4
Evaluate the given integral. $$\int \frac{x}{x^{2}-5 x+6} d x$$
Short Answer
Expert verified
The integral evaluates to: \(-2 \ln|x-2| + 3 \ln|x-3| + C\).
Step by step solution
01
Simplify the Denominator
Firstly, we will factor the quadratic expression in the denominator: \(x^2 - 5x + 6\). This can be factored into \((x-2)(x-3)\). Thus, we can rewrite the integral as \(\int \frac{x}{(x-2)(x-3)} \, dx\).
02
Set Up Partial Fraction Decomposition
The next step is to decompose \(\frac{x}{(x-2)(x-3)}\) into partial fractions. Assume \(\frac{x}{(x-2)(x-3)} = \frac{A}{x-2} + \frac{B}{x-3}\), where \(A\) and \(B\) are constants to be determined.
03
Solve for Constants
Multiply through by the common denominator \((x-2)(x-3)\) to obtain the equation: \(x = A(x-3) + B(x-2)\). Expanding and combining like terms gives: \(x = (A+B)x - (3A + 2B)\). Equate coefficients with \(x = 1x + 0\) to find \(A + B = 1\) and \(-3A - 2B = 0\). Solve these linear equations to get \(A = -2\) and \(B = 3\).
04
Integrate Each Term
With \(A\) and \(B\) found, rewrite the original integral as \(\int \left( \frac{-2}{x-2} + \frac{3}{x-3} \right) \, dx\). This can be evaluated term by term:\(-2 \int \frac{1}{x-2} \, dx + 3 \int \frac{1}{x-3} \, dx\).These integrals are standard, yielding:\(-2 \ln|x-2| + 3 \ln|x-3| + C\), where \(C\) is the constant of integration.
05
Provide the Final Answer
Thus, the evaluated integral is:\(-2 \ln|x-2| + 3 \ln|x-3| + C\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Fraction Decomposition
Partial fraction decomposition is a method used in integral calculus to break down complex rational expressions into simpler ones. This technique is particularly useful when you encounter an integral with a polynomial in the denominator that can be factored into linear factors, like in our given problem.
- Purpose: The main goal is to express a single complex fraction as a sum of simpler ones. Each simpler fraction has a constant numerator and a denominator that is one of the factors of the original denominator.
- Simple form: If we have a fraction like \( \frac{x}{(x-2)(x-3)} \), we assume it can be decomposed into \( \frac{A}{x-2} + \frac{B}{x-3} \).
Indefinite Integration
Indefinite integration is the process of finding a function whose derivative is the given function, plus a constant \(C\) (the "constant of integration"). It's denoted by the integral sign \(\int f(x) \,dx\) and is the reverse process of differentiation.
- Constant of Integration: Since differentiation of a constant is zero, indefinite integrals include an arbitrary constant \(C\). This accounts for all potential antiderivatives.
- Application: Once you have the function in a form amenable to integration, you integrate term by term. In our exercise, after partial fraction decomposition, each term is a standard integral form.
Factorization of Quadratics
Factorization is a mathematical process by which a quadratic polynomial is broken down into a product of its linear factors. This means expressing the quadratic in the form \((x-a)(x-b)\) where \(a\) and \(b\) are the roots of the quadratic.
- Why factorize? It simplifies the integration process by breaking the expression into separate distinct parts, each simpler to handle on its own.
- Methods: A common method for factorization involves finding two numbers that multiply to the constant term and add to the coefficient of the linear term. For example, the quadratic \(x^2 - 5x + 6\) factors into \((x-2)(x-3)\) because \(-2\) and \(-3\) add to \(-5\) and multiply to \(6\).
