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Magazine Chapter 5: Problem 28
In Exercises \(5-28\) , find the indefinite integral. $$\int \frac{x(x-2)}{(x-1)^{3}} d x$$
Short Answer
Expert verified
\(\ln |x - 1| - \frac{1}{2}(x - 1)^{-2} + C\)
Step by step solution
01
Perform substitution
Make a substitution where \( u = x - 1 \), and consequently \( d u = d x \). This simplifies the denominator to \( u^{3} \). Replace all instances of \( x \) in the integral to be in terms of \( u \).
02
Substitute x in terms of u
Express \( x \) in terms of \( u \) to simplify the numerator. From \( u = x - 1 \) , we get \( x = u + 1 \). Substitute \( x = u + 1 \) and \( d x = d u \) into the integral. The integral now becomes \( \int \frac{(u+1)(u+1-2)}{u^{3}} d u = \int \frac{(u+1)(u-1)}{u^{3}} d u\).
03
Split into simpler fractions
Split the fraction under the integral sign into simpler parts. This is done by dividing each term in the numerator by the denominator \( u^{3} \). So the integral becomes \( \int \frac{u^{2}}{u^{3}} - \frac{1}{u^{3}} d u = \int u^{-1} - u^{-3} d u\).
04
Integrate
Now integrate each term separately. For \( u^{-1} \), the anti-derivative is \( \ln |u| \). For \( -u^{-3} \), the antiderivative is \( \frac{1}{2}u^{-2} \) . Therefore, the integral becomes \( \ln |u| - \frac{1}{2}u^{-2} + C \), where \( C \) is the constant of integration.
05
Back-substitute
Finally, back substitute \( u = x - 1 \) to get the integral in terms of the original variable \( x \), so we get \( \ln |x - 1| - \frac{1}{2}(x - 1)^{-2} + C \). This is the indefinite integral of the given function.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Substitution Method
The substitution method is a powerful tool in calculus used to simplify the process of finding integrals. Think of it as a way to "change the variable" to make integration more manageable. Here's how it works:
- The main idea is to choose a substitution to simplify the integral. This often involves setting a part of the integral, usually a complicated expression, equal to a new variable, often denoted as \( u \).
- For example, in the provided exercise, we selected \( u = x - 1 \), which helps transform the original complex expression into a simpler form. This choice is strategic to cancel out complexities and make the integral solvable.
- The substitution also requires us to express \( dx \) in terms of \( du \) for the integral to be calculated effectively. In this case, since \( du = dx \), the substitution doesn't change the differential, making calculations smoother.
Antiderivative
An antiderivative is essentially a function that reverses the process of differentiation. When we're given a derivative, finding its antiderivative means answering the question: "What function was differentiated?" In the context of the indefinite integral, the antiderivative represents the solution of the integral.
- As we processed the given integral, \( \int u^{-1} - u^{-3} \, du \), we found two antiderivatives in simpler terms.
- For \( u^{-1} \), the antiderivative is \( \ln |u| \). This comes from the rule that integrates \( u^{-1} \) to the natural logarithm of \( u \).
- For \( -u^{-3} \), it became \( \frac{1}{2}u^{-2} \). Applying the power rule of integration helps find the expression that results when the derivative was originally taken.
Back-substitution
Back-substitution is a crucial step that follows the substitution method. Once we have evaluated the integral with the substitution, this process helps revert back to the original variable.
- In our exercise, after solving the integral in terms of \( u \), we need to replace \( u \) back with the original variable \( x \).
- Back-substitution involved replacing \( u = x - 1 \). Therefore, every \( u \) present in our antiderivative could now be rewritten as \( x - 1 \). This yields the result \( \ln |x - 1| - \frac{1}{2}(x - 1)^{-2} + C \).
Constant of Integration
The constant of integration, often denoted by \( C \), is an essential component in the realm of indefinite integrals. Understanding its role enriches your grasp of fundamental calculus.
- When we integrate, we're performing the reverse of differentiation, which means we need to account for what might have been "lost"鈥攁ny constant terms, since derivatives of constants are zero.
- The constant \( C \) thus represents an infinite number of possible solutions. Every distinct value of \( C \) makes a different but valid antiderivative, representing a whole family of functions.
- In the solution obtained, we see \( C \) added to the expression \( \ln |x - 1| - \frac{1}{2}(x - 1)^{-2} + C \). This ensures the solution is complete, embracing all possible original functions whose derivative is the given integral.
