Chapter 30: Problem 2
\(2 \int \frac{d x}{8-12 x-x^{2}}\)
Short Answer
Expert verified
The integral evaluates to \(2 \ln \left| \frac{x+2}{x+4} \right| + C\).
Step by step solution
01
Simplify the Integral
To solve the integral \(2 \int \frac{d x}{8-12 x-x^{2}}\), we first simplify the expression in the denominator. Notice it resembles a quadratic expression. We can factor it to make the integration process simpler.
02
Factor the Quadratic Expression
Rewrite the quadratic expression \(-x^2 - 12x + 8\). For easier factoring, start by rewriting it as \(-1(x^2 + 12x - 8)\). By checking for factors of \(c = -8\) that satisfy the sum \(b = 12\), we factor it as \(-1(x + 2)(x + 4)\). So, the integral becomes \(2 \int \frac{d x}{-1(x + 2)(x + 4)}\).
03
Rewrite the Integral with Partial Fraction Decomposition
Express \(\frac{1}{(x+2)(x+4)}\) as a partial fraction: \(\frac{A}{x+2} + \frac{B}{x+4}\). Solve by setting equal to: \[ 1 = A(x+4) + B(x+2) \] Expand and Equate coefficients: - \(A + B = 0\)- \(4A + 2B = 1\)Solving gives \(A = \frac{1}{2}, B = -\frac{1}{2}\).
04
Integrate the Expression
Now substitute back the partial fractions into the integral \[ 2 \int \left( \frac{1/2}{x+2} - \frac{1/2}{x+4} \right) d x = \int \frac{1}{x+2} d x - \int \frac{1}{x+4} d x \]The integrals evaluate to \[ \ln|x+2| - \ln|x+4| + C \].
05
Simplify the Result
Using the logarithmic property \(\ln a - \ln b = \ln \left(\frac{a}{b}\right)\), simplify the expression to:\[\ln \left| \frac{x+2}{x+4} \right| + C \]
06
Write the Final Answer
The solution to the integral \(2 \int \frac{d x}{8-12 x-x^{2}}\) is \[ 2 \ln \left| \frac{x+2}{x+4} \right| + C \], where \(C\) is the constant of integration.
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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 technique that helps in simplifying complex fractions into simpler ones, making them easier to integrate. This is particularly useful when you have a rational expression, like the one in our exercise, where the denominator is a factored polynomial. Understanding the basics of partial fraction decomposition involves breaking down a fraction into a sum of fractions.
To apply this technique, the denominator must be fully factored. For example, with the integral \( \frac{1}{(x+2)(x+4)} \), we can express it as \( \frac{A}{x+2} + \frac{B}{x+4} \). The goal is to determine the values of \( A \) and \( B \) that satisfy the condition when the fractions are combined back.
Here's how you typically go about it:
To apply this technique, the denominator must be fully factored. For example, with the integral \( \frac{1}{(x+2)(x+4)} \), we can express it as \( \frac{A}{x+2} + \frac{B}{x+4} \). The goal is to determine the values of \( A \) and \( B \) that satisfy the condition when the fractions are combined back.
Here's how you typically go about it:
- Multiply both sides by the original denominator to get rid of the fraction.
- Expand the right side and collect like terms.
- Equate coefficients of corresponding powers of \( x \) and solve the resulting system of equations.
Quadratic Expression Factorization
Quadratic expression factorization is crucial when dealing with integrals where the denominator is a quadratic polynomial. Factoring transforms the quadratic expression into a product of simpler factors, which can then be handled separately. In the given problem, the quadratic expression was \(-x^2 - 12x + 8\).
We can factor it by first rewriting as \(-1(x^2 + 12x - 8)\). The next step is to find factors of the constant term \(-8\) that add up to the linear coefficient, which is \(12\). After examining possible combinations, we realize it factors into \(-(x+2)(x+4)\).
Key points for factorization:
We can factor it by first rewriting as \(-1(x^2 + 12x - 8)\). The next step is to find factors of the constant term \(-8\) that add up to the linear coefficient, which is \(12\). After examining possible combinations, we realize it factors into \(-(x+2)(x+4)\).
Key points for factorization:
- Rewriting the quadratic in standard form \(ax^2 + bx + c\).
- Searching for two numbers that multiply to \(ac\) and sum to \(b\).
- Breaking the middle term if necessary and factoring by grouping.
Logarithmic Properties
Logarithmic properties become quite handy, especially when dealing with integrals of the form \(\frac{1}{x}\), which integrate to logarithmic functions. After integrating, we simplify further using properties of logarithms, which are very powerful for manipulating and simplifying expressions.
Consider the integrated expression: \[\ln|x+2| - \ln|x+4| + C\].
The logarithmic property \(\ln a - \ln b = \ln \left( \frac{a}{b} \right)\) allows us to combine the two logarithms into \[\ln \left| \frac{x+2}{x+4} \right| + C\].
Benefits of understanding logarithmic properties:
Consider the integrated expression: \[\ln|x+2| - \ln|x+4| + C\].
The logarithmic property \(\ln a - \ln b = \ln \left( \frac{a}{b} \right)\) allows us to combine the two logarithms into \[\ln \left| \frac{x+2}{x+4} \right| + C\].
Benefits of understanding logarithmic properties:
- Turns subtraction of logs into division, simplifying expressions.
- Consolidates expressions, making them neater and often easier to interpret.
- Helps in solving equations where logarithmic terms are present.