Chapter 5: Problem 43
Find each indefinite integral by the substitution method or state that it cannot be found by our substitution formulas. \(\int \frac{\ln x}{x} d x \quad[\) Hint \(:\) Let \(u=\ln x .]\)
Short Answer
Expert verified
The integral is \( \frac{(\ln{x})^2}{2} + C \).
Step by step solution
01
Identify the Substitution
Given the hint, we choose the substitution \( u = \ln{x} \). This implies that the derivative of \( u \) with respect to \( x \) is \( \frac{du}{dx} = \frac{1}{x} \).
02
Rearrange for Differential
From \( \frac{du}{dx} = \frac{1}{x} \), we rearrange for \( dx \) to obtain \( du = \frac{1}{x} dx \) or \( dx = x \, du \).
03
Substitute in the Integral
Substitute \( u = \ln{x} \) and \( dx = x \cdot du \) back into the integral. Since \( \frac{dx}{x} = du \), the integral becomes \( \int u \, du \).
04
Integrate with Respect to u
The integral \( \int u \, du \) is a standard power rule integral. The antiderivative is \( \frac{u^2}{2} + C \), where \( C \) is the constant of integration.
05
Substitute Back to Original Variable
Replace \( u \) with \( \ln{x} \) to return to the original variable: \( \frac{(\ln{x})^2}{2} + C \). Thus, the indefinite integral is \( \frac{(\ln{x})^2}{2} + C \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Substitution Method in Integration
The substitution method is a powerful technique used in integral calculus to simplify finding antiderivatives. When we have a complex integral, sometimes substituting a part of the function with a new variable makes the integration process more manageable. In our exercise, we use the hint to let \( u = \ln x \). This simplifies the integral and allows us to deal with a less complicated expression.
- Step 1 involves choosing a substitution that captures the complexity of the integral. Here, the derivative of \( \ln x \) is simple, \( \frac{1}{x} \), making it a prime candidate for substitution.
- After substitution, the integral's variable is changed from \( x \) to \( u \), transforming the integral into a form that's easier to solve.
Understanding Derivatives
A derivative represents the rate of change of a function. In the context of integration, derivatives help us choose appropriate substitutions and solve for differentials. In our example, noticing that the derivative of \( \ln x \) is \( \frac{1}{x} \) is crucial.
- This insight allows us to express \( dx \) in terms of \( du \) as \( dx = x \, du \).
- This step is vital because it transforms the integrand into a function of \( u \), enabling simpler integration.
Antiderivative: Reversing Derivatives
An antiderivative is essentially the opposite of a derivative. It is a function whose derivative gives back the function from which we started. Finding an antiderivative is the critical step in determining the indefinite integral.In our exercise, after substituting \( u = \ln x \), we are left with the integral \( \int u \, du \). This involves using the power rule for integration: when you integrate \( u^n \), you increase the power by one and divide by the new power.
- The antiderivative of \( u \) is calculated as \( \frac{u^2}{2} + C \), where \( C \) is the constant of integration.
- This formulaic step is integral to finding the final answer by ensuring your integrated function differentiates back to the original form.
Integral Calculus: The Indefinite Integral
Integral calculus is the branch of mathematics dealing with integrals and their applications. The indefinite integral, unlike the definite integral which calculates an area, results in a general formula for the antiderivative of a function. In our exercise, we started with \( \int \frac{\ln x}{x} \, dx \) which we solved using a substitution technique. After finding the antiderivative in terms of \( u \), we substitute back the original variable to get the indefinite integral: \( \frac{(\ln x)^2}{2} + C \).
- This indefinite integral expression represents all possible antiderivatives of the given function.
- The "+ C" is significant because it accounts for any constant that could have been part of the original function but disappears upon differentiation.