Problem 33
Question
In Problems 17-36, use substitution to evaluate each indefinite integral. $$ \int \frac{(\ln x)^{2}}{x} d x $$
Step-by-Step Solution
Verified Answer
The integral evaluates to \( \frac{(\ln x)^3}{3} + C \).
1Step 1: Identify the substitution
For the given integral \( \int \frac{(\ln x)^{2}}{x} \, dx \), notice that the derivative of \( \ln x \) is \( \frac{1}{x} \). This suggests substituting \( u = \ln x \).
2Step 2: Differentiate the substitution
Differentiate \( u = \ln x \) to express \( dx \) in terms of \( du \). The derivative is \( \frac{du}{dx} = \frac{1}{x} \), so we have \( du = \frac{1}{x} \, dx \). This implies that \( dx = x \, du \), but since \( x \, dx = du \), we get simply \( dx = x \, du \).
3Step 3: Substitute into the integral
Substitute \( u = \ln x \) and \( dx = x \, du \) back into the integral. The integral becomes \( \int u^{2} \, du \).
4Step 4: Integrate with respect to \( u \)
Integrate \( \int u^{2} \, du \) using the power rule. The integral is \( \frac{u^3}{3} + C \) where \( C \) is the constant of integration.
5Step 5: Back-substitute \( u \)
Replace \( u \) with \( \ln x \) to express the integral in terms of \( x \). Thus, the solution becomes \( \frac{(\ln x)^3}{3} + C \).
Key Concepts
Substitution MethodIntegration by PartsPower Rule
Substitution Method
The substitution method is a technique used for simplifying complex integrals. It involves replacing a function inside the integral with a new variable, making the integral easier to solve. This method is akin to the substitution in algebra, where variables are swapped to solve easier equations.
To use this method, first identify a part of the integrand (the expression inside the integral) whose derivative is also present. In our example, we noticed that the derivative of \( \ln x \) is \( \frac{1}{x} \). This is our cue to substitute.
To use this method, first identify a part of the integrand (the expression inside the integral) whose derivative is also present. In our example, we noticed that the derivative of \( \ln x \) is \( \frac{1}{x} \). This is our cue to substitute.
- **Pick a suitable substitution:** In this case, set \( u = \ln x \).
- **Differentiate your substitution:** This gives \( \frac{du}{dx} = \frac{1}{x} \), so rearranging terms, \( du = \frac{1}{x} \, dx \).
- **Replace in the integral:** Substitute these into the original integral, transforming \( \int \frac{(\ln x)^{2}}{x} \, dx \) to \( \int u^{2} \, du \).
Integration by Parts
Integration by parts is a technique derived from the product rule of differentiation. It's particularly useful when your integral involves a product of two functions. Though not directly used in our current solution, it's another indispensable tool in the calculus toolbox.
The formula for integration by parts is:
\[\int u \, dv = uv - \int v \, du\]
This involves choosing parts of the integrand as \(u\) and \(dv\), then differentiating and integrating these to find \(du\) and \(v\), respectively.
For example, if you have an integral involving \(x \cdot e^x\), you might let \(u = x\) (for which \(du = dx\)) and \(dv = e^x \, dx\), allowing the technique to simplify the problem.
It's important to choose \(u\) and \(dv\) wisely to ensure that \(\int v \, du\) is simpler than the initial integral. Often, polynomial terms (like \(x\)) are chosen as \(u\) because they become simpler when differentiated.
The formula for integration by parts is:
\[\int u \, dv = uv - \int v \, du\]
This involves choosing parts of the integrand as \(u\) and \(dv\), then differentiating and integrating these to find \(du\) and \(v\), respectively.
For example, if you have an integral involving \(x \cdot e^x\), you might let \(u = x\) (for which \(du = dx\)) and \(dv = e^x \, dx\), allowing the technique to simplify the problem.
It's important to choose \(u\) and \(dv\) wisely to ensure that \(\int v \, du\) is simpler than the initial integral. Often, polynomial terms (like \(x\)) are chosen as \(u\) because they become simpler when differentiated.
Power Rule
The power rule is a fundamental technique in calculus for finding the integral of power functions. It’s incredibly straightforward and essential for solving polynomial integrals.
The rule states:
\[\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\]
Here, \(n\) is any real number except \(-1\) since the integral of \( \frac{1}{x} \) is \( \ln |x| + C \).
In our example, after substitution, we ended up with \( \int u^2 \, du \). The power rule is directly applicable here:
The rule states:
\[\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\]
Here, \(n\) is any real number except \(-1\) since the integral of \( \frac{1}{x} \) is \( \ln |x| + C \).
In our example, after substitution, we ended up with \( \int u^2 \, du \). The power rule is directly applicable here:
- **Apply the Power Rule:** Increasing the power by one, \( 2+1 = 3 \), and dividing by the new power, the integral becomes \( \frac{u^3}{3} + C \).
Other exercises in this chapter
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