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Evaluate the following integrals. $$\int \frac{8}{(x-2)(x+6)} d x$$

Short Answer

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Question: Determine the integral of the function \(\frac{8}{(x-2)(x+6)}dx\). Solution: The integral of the given function is \(\int \frac{8}{(x-2)(x+6)} dx = \ln\Big|\frac{x-2}{x+6}\Big|+C\), where C is the constant of integration.

Step by step solution

01

Setup Partial Fraction Decomposition

We need to rewrite the fraction \(\frac{8}{(x-2)(x+6)}\) as the sum of two simpler fractions, where the denominators are \((x-2)\) and \((x+6)\), respectively. $$\frac{8}{(x-2)(x+6)} = \frac{A}{(x-2)} + \frac{B}{(x+6)}$$
02

Solve for A and B

To solve for A and B, clear the denominators by multiplying both sides of the equation by the common denominator \((x-2)(x+6)\): $$8=\frac{A(x-2)(x+6)+B(x-2)(x+6)}{(x-2)(x+6)}$$ $$8=A(x+6)+B(x-2)$$ Now, we can use two values of x to solve for A and B. We will use x=2 and x=-6. For x=2: $$8=A(2+6)$$ $$8=8A$$ So, \(A=1\). For x=-6: $$8=B(-6-2)$$ $$8=-8B$$ So, \(B=-1\). Now we have our partial fraction decomposition: $$\frac{8}{(x-2)(x+6)} = \frac{1}{(x-2)} - \frac{1}{(x+6)}$$
03

Integrate Each Fraction

Now, we can integrate each fraction separately: $$\int \frac{8}{(x-2)(x+6)} d x= \int\Bigg(\frac{1}{(x-2)} - \frac{1}{(x+6)}\Bigg) d x$$ $$=\int\frac{1}{(x-2)} dx - \int\frac{1}{(x+6)} dx$$
04

Apply Integration Formulas and Simplify

Using the natural logarithm integration formula, we have: $$=\ln| x -2 |- \ln |x+6| + C$$ $$=\ln\Big|\frac{x-2}{x+6}\Big|+C$$ So, the final solution for the integral is: $$\int \frac{8}{(x-2)(x+6)} d x =\ln\Big|\frac{x-2}{x+6}\Big|+C$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integral Calculus
Integral Calculus is a branch of mathematics that deals with the concept of integration. Integration is the reverse process of differentiation. It helps us find the whole from its parts, which is often represented by the area under a curve. In this particular example, we are evaluating the integral of the function \( \frac{8}{(x-2)(x+6)} \).

The integral represents the accumulation of quantities, and in calculus, it is typically used to find areas, volumes, central points, or other useful things. There are two main types of integrals: definite and indefinite. In the context of this exercise, we are dealing with an indefinite integral because no limits of integration are provided. This means the result will include an arbitrary constant, \( C \), representing all possible vertical shifts of the original function.

By using the method of partial fractions, we decompose the complex rational expression into simpler fractions that are easier to integrate. This step simplifies the process of integration, allowing us to leverage basic integral calculus rules and derive the final expression neatly.
Natural Logarithm
The natural logarithm is a crucial concept in calculus, often denoted as \( \ln(x) \). The natural logarithmic function is the inverse of the exponential function \( e^x \). In the context of integration, if you integrate \( \frac{1}{x} \), the result is the natural logarithm of the absolute value of \( x \), symbolically represented as \( \ln|x| \) plus a constant of integration \( C \).

In this exercise, after applying partial fraction decomposition, the result is two simpler functions: \( \frac{1}{x-2} \) and \( -\frac{1}{x+6} \). Integration of these two functions utilizes the logarithmic integration technique, yielding \( \ln|x-2| \) and \( \ln|x+6| \), respectively.

  • The absolute value signs in the natural logarithm function ensure that it is defined for all non-zero real numbers.
  • The integration of these terms follows straightforwardly from the property of derivatives and integrals associated with logarithmic functions.
  • Finally, combining these results gives the simplified natural logarithmic form \( \ln\Big|\frac{x-2}{x+6}\Big| \), representing the expression in logarithmic format after integrating.
Integration Techniques
In calculus, mastering various integration techniques is essential for solving complex integral problems. There are several common methods used to evaluate integrals when standard formulas do not directly apply. In this exercise, we primarily use two techniques: **Partial Fraction Decomposition** and **Logarithmic Integration**.

Partial fraction decomposition is useful when dealing with rational functions. It involves breaking down a complicated fraction into a sum of simpler fractions, each of which is easier to integrate. This method only works if the degree of the numerator is less than that of the denominator. By finding constants \(A\) and \(B\) such that the original expression is rewritten in terms of these constants and simpler fractions, integration becomes more manageable.

Logarithmic integration, on the other hand, applies the rule that the integral of \( \frac{1}{x} \) is \( \ln|x| \). This technique is ideal for functions resulting from partial fraction decomposition because they often resemble the form \( \frac{1}{x-a} \) after decomposition.

By combining these techniques, we can handle complex integrals and systematically break them down into simpler, solvable components. This step-by-step approach ensures clarity and understanding in calculus problems, leading to the solution \( \ln\Big|\frac{x-2}{x+6}\Big|+C \) for the given integral.

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