Chapter 28: Problem 33
Solve the given problems by integration.Integrate \(\int \frac{x-4}{x+4} d x\) by first using algebraic division to change the form of the integrand.
Short Answer
Expert verified
The integral is \(x - 8 \ln |x+4| + C\).
Step by step solution
01
Perform Algebraic Division
Divide the numerator \(x-4\) by the denominator \(x+4\). By performing algebraic division, \(x-4\) divided by \(x+4\) gives us \(1 - \frac{8}{x+4}\). This means that the integral becomes \(\int (1 - \frac{8}{x+4}) \, dx\).
02
Integrate Each Term Separately
Now, break the integral \(\int (1 - \frac{8}{x+4}) \, dx\) into two separate integrals: \(\int 1 \, dx\) and \(-8 \int \frac{1}{x+4} \, dx\).
03
Integrate the First Term
Integrate \(\int 1 \, dx\). This integration results in \(x\).
04
Integrate the Second Term
Integrate \(-8 \int \frac{1}{x+4} \, dx\). This integration results in \(-8 \ln |x+4|\).
05
Combine the Results and Add the Constant of Integration
Combine the results from Step 3 and Step 4 to obtain the integrated function. Thus, the solution to the integral is \(x - 8 \ln |x+4| + C\), 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.
Algebraic Division
Understanding algebraic division can simplify the process of integration. It's a technique similar to long division but applied to algebraic expressions. In this exercise, we divided the numerator \(x-4\) by the denominator \(x+4\). To do this, you start by dividing the first term of the numerator \(x\) by the first term of the denominator \(x\), giving you 1 as a result. This 1 multiplies the entire divisor \(x+4\), subtracting from \(x-4\) and leaving a remainder of -8. Thus, the entire expression \(\frac{x-4}{x+4}\) can be rewritten as \(1 - \frac{8}{x+4}\).
This division helps:
This division helps:
- Simplify complex fractions
- Make integration more straightforward
- Expose any hidden term that might be integrated independently
Definite Integrals
Definite integrals involve finding the area under a curve for a specified interval. However, in this exercise, our focus is on indefinite integrals which emphasize the anti-derivative without limits. Even though definite integrals aren't specifically used here, understanding them helps to appreciate where we might apply the integration skills.
In definite integrals:
In definite integrals:
- You compute the net area under the curve between two points.
- Boundaries are specified on the integral sign, \(\int^{b}_{a}\).
- The result is a numerical value representing this area.
Natural Logarithm
The natural logarithm, represented by \(\ln\), is the inverse of the exponential function \(e^x\). It's a critical part of the integration in the given exercise, especially when dealing with fractions or rational functions.
After integrating \(-8 \int \frac{1}{x+4} \, dx\), the result is \(-8 \ln |x+4|\), which is derived from the integral of \(\frac{1}{x+a}\). Here's where the properties of natural logarithms shine:
After integrating \(-8 \int \frac{1}{x+4} \, dx\), the result is \(-8 \ln |x+4|\), which is derived from the integral of \(\frac{1}{x+a}\). Here's where the properties of natural logarithms shine:
- Natural logs simplify expressions, particularly as \(\ln(e^x) = x\).
- They are essential in solving differential equations and providing solutions to growth models.
- The absolute value, \(|x+4|\), ensures the argument remains positive, which is necessary since the domain of \(\ln(x)\) is \(x > 0\).