Problem 43
Question
Use partial fraction decompositions to evaluate each integral. $$ \int \frac{x-2}{(x-1)^{2}} d x $$
Step-by-Step Solution
Verified Answer
The integral is \( \ln|x-1| - \frac{1}{x-1} + C \).
1Step 1: Setup the Partial Fraction Decomposition
To decompose the function \( \frac{x-2}{(x-1)^2} \) into partial fractions, we express it as \( \frac{A}{x-1} + \frac{B}{(x-1)^2} \). The goal is to find constants \( A \) and \( B \) that satisfy this equality.
2Step 2: Clear the Denominator and Simplify
Multiply both sides by \((x-1)^2\) to clear the denominator: \( x-2 = A(x-1) + B \). Then expand the right side: \( x-2 = Ax - A + B \).
3Step 3: Solve for Coefficients A and B
By comparing coefficients on both sides of \( x-2 = Ax - A + B \), we can derive the following equations: \( A = 1 \) and \( -A + B = -2 \). Solving these, \( A = 1 \) and \( B = -1 \).
4Step 4: Substitute Back into the Partial Fractions
Using the determined values \( A = 1 \) and \( B = -1 \), substitute back: \( \frac{1}{x-1} + \frac{-1}{(x-1)^2} \). We will integrate each of these terms separately.
5Step 5: Integrate Each Fraction Separately
Now integrate each term separately: \[ \int \frac{1}{x-1} \, dx + \int \frac{-1}{(x-1)^2} \, dx. \]The first integral results in \( \ln|x-1| \), and the second integral becomes \( \frac{1}{x-1} \) as \( \int \frac{1}{u^2} du = -\frac{1}{u}\) where \( u = x-1 \).
6Step 6: Combine the Results
Combine the results from each integral to get the final answer: \( \ln|x-1| - \frac{1}{x-1} + C \), where \( C \) is the integration constant.
Key Concepts
Partial Fraction DecompositionIntegration TechniquesIntegral Calculus
Partial Fraction Decomposition
Partial Fraction Decomposition is a method used to break down complex rational expressions into simpler fractions that are easier to integrate. Imagine you have a complicated fraction like \( \frac{x-2}{(x-1)^2} \). It could seem daunting at first, but by using partial fraction decomposition, we can simplify it into a sum of fractions. This technique is incredibly helpful in calculus, especially when dealing with integration problems.
The goal is to express \( \frac{x-2}{(x-1)^2} \) as a sum of two simpler fractions: \( \frac{A}{x-1} + \frac{B}{(x-1)^2} \). Here, \( A \) and \( B \) are constants that we need to find.
Firstly, after setting up the equation, you multiply both sides by the denominator \((x-1)^2\) to eliminate the fraction. What you have left is an equation without denominators: \( x-2 = A(x-1) + B \). By expanding and comparing coefficients of \(x\) and constants, you can solve for \(A\) and \( B\). In this case, \( A = 1 \) and \( B = -1 \).
The goal is to express \( \frac{x-2}{(x-1)^2} \) as a sum of two simpler fractions: \( \frac{A}{x-1} + \frac{B}{(x-1)^2} \). Here, \( A \) and \( B \) are constants that we need to find.
Firstly, after setting up the equation, you multiply both sides by the denominator \((x-1)^2\) to eliminate the fraction. What you have left is an equation without denominators: \( x-2 = A(x-1) + B \). By expanding and comparing coefficients of \(x\) and constants, you can solve for \(A\) and \( B\). In this case, \( A = 1 \) and \( B = -1 \).
Integration Techniques
Integration techniques are strategies used to solve integrals, which involve finding the reverse of differentiation. With our expression \( \frac{1}{x-1} + \frac{-1}{(x-1)^2} \), standard integration techniques are applied to tackle each term individually.
For the first term, \( \frac{1}{x-1} \), the technique used is simple: it's a basic form that integrates to the natural logarithm, specifically \( \ln|x-1| \). This is an essential integration result you should remember.
For the second term, \( \frac{-1}{(x-1)^2} \), we identify it as a power function. The integral \( \int \frac{1}{u^2} du \) follows a form that integrates to \( -\frac{1}{u} \), where \( u \) is \( x-1 \) in this scenario. By practicing such techniques, you can tackle myriad types of integrands you will encounter.
For the first term, \( \frac{1}{x-1} \), the technique used is simple: it's a basic form that integrates to the natural logarithm, specifically \( \ln|x-1| \). This is an essential integration result you should remember.
For the second term, \( \frac{-1}{(x-1)^2} \), we identify it as a power function. The integral \( \int \frac{1}{u^2} du \) follows a form that integrates to \( -\frac{1}{u} \), where \( u \) is \( x-1 \) in this scenario. By practicing such techniques, you can tackle myriad types of integrands you will encounter.
Integral Calculus
Integral Calculus is the branch of calculus that focuses on finding the function that describes the accumulation of quantities. When we're working with integration, we're essentially piecing together an original function from its derivative.
After decomposing the fraction and integrating each term separately, we combined the results. We integrate to find:
Combining these results gives us the solution to the original integral. Don’t forget the constant \( C \), as integration involves finding the antiderivative and allows for any constant shift.
In essence, integral calculus helps us reverse the process of differentiation, allowing us to understand and quantify areas under curves and solve real-world problems involving rates and accumulations of change.
After decomposing the fraction and integrating each term separately, we combined the results. We integrate to find:
- \( \ln|x-1| \) for the \( \frac{1}{x-1} \) part
- \( \frac{1}{x-1} \) for the \( \frac{-1}{(x-1)^2} \)
Combining these results gives us the solution to the original integral. Don’t forget the constant \( C \), as integration involves finding the antiderivative and allows for any constant shift.
In essence, integral calculus helps us reverse the process of differentiation, allowing us to understand and quantify areas under curves and solve real-world problems involving rates and accumulations of change.
Other exercises in this chapter
Problem 43
Find a comparison function for each integrand and determine whether the integral is convergent. $$ \int_{1}^{\infty} \frac{1}{\sqrt{1+x}} d x $$
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In Problems \(39-48\), first make an appropriate substitution and then use integration by parts to evaluate each integral. $$ \int x^{3} \sin \left(x^{2}\right)
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In Problems \(43-58\), use substitution to evaluate each definite integral. $$ \int_{0}^{3} x \sqrt{x^{2}+1} d x $$
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Find a comparison function for each integrand and determine whether the integral is convergent. $$ \int_{0}^{\infty} \frac{1}{e^{x}+e^{-x}} d x $$
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