/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 14 Evaluate the following integrals... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 14

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

Short Answer

Expert verified
Based on the given step by step solution, create a short answer: To evaluate the integral of the given rational function, $\int \frac{8}{(x-2)(x+6)} dx$, first perform partial fraction decomposition. You will end up with the following expression: $\frac{1}{x-2} - \frac{1}{x+6}$. Now, integrate term by term: $$\int \frac{8}{(x-2)(x+6)} dx = \ln\left|\frac{x-2}{x+6}\right| + C$$, where \(C\) is the constant of integration.

Step by step solution

01

Find the partial fraction decomposition of the integrand

First, express the rational function as the sum of its partial fractions. To do this, we follow these steps: 1. Factoring the denominator: there is no need to factor it further since it's already given as \((x-2)(x+6)\). 2. Writing the general form of the partial fractions decomposition: For each factor, we write the term and assign a constant as a coefficient. In this case, we have: $$\frac{8}{(x-2)(x+6)} = \frac{A}{x-2} + \frac{B}{x+6}$$
02

Solve for A and B

Now, we move on to solve for constants \(A\) and \(B\). To do this, we clear the denominators and equate the numerators: $$8 = A(x+6) + B(x-2)$$ Now we can solve for \(A\) and \(B\) by employing two different methods: Method 1: Pick convenient values for x that eliminate one of the constants. In our case, let's pick \(x = 2\) and \(x = -6\). This gives: $$8 = A(2+6) \Rightarrow A = 1$$ $$8 = B(-6-2) \Rightarrow B = -1$$ Method 2: Expanding, equating coefficients, and solving the system of equations: $$8 = (A+B)x + 6A - 2B$$ This gives the system of equations: $$A+B=0$$ $$6A - 2B = 8$$ Solving this system, we also find that \(A = 1\) and \(B = -1.\)
03

Integrate term by term

Now that we have the partial fractions decomposition as: $$\frac{8}{(x-2)(x+6)} = \frac{1}{x-2} - \frac{1}{x+6}$$ We can integrate term by term: $$\int \frac{8}{(x-2)(x+6)} dx = \int \left(\frac{1}{x-2} - \frac{1}{x+6}\right) dx$$ Applying the rules of antiderivatives, we obtain: $$\int \frac{8}{(x-2)(x+6)} dx = \ln |x-2| - \ln |x+6| + C$$ where \(C\) is the constant of integration. Finally, we can combine the logarithms by using the logarithmic property \(\ln{a}-\ln{b}=\ln{\frac{a}{b}}\) to obtain the final answer: $$\int \frac{8}{(x-2)(x+6)} dx = \ln\left|\frac{x-2}{x+6}\right| + C$$

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