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Chapter 8: Problem 36

Evaluate the integrals. Some integrals do not require integration by parts. $$ \int \frac{(\ln x)^{3}}{x} d x $$

Short Answer

Expert verified
\( \int \frac{(\ln x)^{3}}{x} dx = \frac{(\ln x)^4}{4} + C \).

Step by step solution

01

Identify the integral and decide on the method

We have the integral \( \int \frac{(\ln x)^{3}}{x} d x \). We notice that the integrand is a power of the natural logarithm divided by \( x \). A common method to tackle such integrals involves a substitution where \( u = \ln x \). This simplifies the integral considerably.
02

Perform the substitution

Let \( u = \ln x \). Then, the derivative \( du = \frac{1}{x} dx \), which implies that \( dx = x \, du \). Given that \( x = e^u \) (since \( u = \ln x \)), our integral transforms as follows:\[\int \frac{(\ln x)^{3}}{x} d x = \int u^3 \, du.\]
03

Integrate the new function

Now we integrate \( \int u^3 \, du \). Using the power rule for integration, we find:\[\int u^3 \, du = \frac{u^4}{4} + C.\]
04

Substitute back the original variable

We substitute \( u = \ln x \) back into the integrated result:\[\frac{(\ln x)^4}{4} + C.\]
05

Verify the solution

Differentiate \( \frac{(\ln x)^4}{4} + C \) with respect to \( x \):\[\frac{d}{dx}\left( \frac{(\ln x)^4}{4} \right) = \frac{1}{4} \cdot 4(\ln x)^3 \cdot \frac{1}{x} = \frac{(\ln x)^3}{x}.\]This confirms that the antiderivative was correctly found.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Natural Logarithm
The natural logarithm, often denoted as \( \ln x \), is a special mathematical function that is crucial in calculus and many other areas of mathematics. It is the inverse function of the exponential function with base \( e \), where \( e \) is approximately 2.71828. The function \( \ln x \) plays a prominent role in integration, especially when dealing with problems involving growth or decay models.

Some key properties of the natural logarithm include:
  • \( \ln(1) = 0 \) because \( e^0 = 1 \).
  • \( \ln(e) = 1 \) since \( e^1 = e \).
  • \( \ln(xy) = \ln(x) + \ln(y) \), indicating that the natural logarithm of a product is the sum of the logarithms.
  • \( \ln\left(\frac{x}{y}\right) = \ln(x) - \ln(y) \).
Understanding these properties can help simplify complex integrals, such as the integral \( \int \frac{(\ln x)^3}{x} dx \) by recognizing that \( \ln x \) can be treated as a variable for substitution, leading to an easier problem.
Power Rule
The power rule for integration is a fundamental concept in calculus used to find antiderivatives. It states that if \( n eq -1 \), the antiderivative of \( x^n \) is:
\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \]
Where \( C \) is the constant of integration.

In the context of the provided integral, \( \int u^3 \, du \), the power rule is applied. By identifying that \( u = \ln x \), the power rule simplifies the integration process:
  • The power \( n \) in this case is 3.
  • By substituting into the formula, we integrate to get \( \frac{u^4}{4} + C \).
The power rule effectively reduces the complexity of polynomial expressions, transforming them into simpler, solvable integrals.
Antiderivative
Finding an antiderivative or an indefinite integral involves determining a function whose derivative matches the given function. It is essentially the reverse of differentiation, and its result includes a constant of integration \( C \).

In our exercise, the function \( u^3 \) where \( u = \ln x \) needed integration. The antiderivative of \( u^3 \) is \( \frac{u^4}{4} + C \).

This process confirms the integral's correctness by taking the derivative of the antiderivative and ensuring the original function is recovered:
  • Derive \( \frac{(\ln x)^4}{4} \), resulting in \( \frac{1}{4} \cdot 4(\ln x)^3 \cdot \frac{1}{x} = \frac{(\ln x)^3}{x} \).
  • This step verifies that the original integrand was accurate.
Antiderivatives are crucial for solving differential equations, calculating areas, and resolving various calculus-related problems.
Definite and Indefinite Integrals
Integrals can be classified into two main types: definite and indefinite. Understanding the difference is key to solving integrals correctly.

Indefinite integrals, like the one in our exercise, represent the collection of all possible antiderivatives of a function. They include an arbitrary constant \( C \) due to the nature of antiderivates:
  • For example, \( \int f(x) \, dx = F(x) + C \) indicates the general form of the antiderivative.
Definite integrals, on the other hand, calculate the accumulated value, or area under the curve, between two points \( a \) and \( b \):
  • Represented as \( \int_{a}^{b} f(x) \, dx \).
  • They do not include the constant \( C \) because they provide a specific numerical result.
Both types rely on finding antiderivatives, but definite integrals further involve evaluating the result at specific bounds to find specific values rather than a general formula.

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Most popular questions from this chapter

Use a CAS to evaluate the integrals. $$\int_{0}^{2 / \pi} x \sin \frac{1}{x} d x$$

In Exercises \(87-90,\) use a CAS to explore the integrals for various values of \(p\) (include noninteger values). For what values of \(p\) does the integral converge? What is the value of the integral when it does converge? Plot the integrand for various values of \(p\) . $$\int_{e}^{\infty} x^{p} \ln x d x$$

Estimating the value of a convergent improper integral whose domain is infinite \begin{equation} \begin{array}{l}{\text { a. Show that }} \\\\\quad\quad\quad\quad\quad\quad {\int_{3}^{\infty} e^{-3 x} d x=\frac{1}{3} e^{-9}<0.000042} \\ \quad{\text { and hence that } \int_{0}^{\infty} e^{-x^{2}} d x<0.000042 . \text { Explain why this }} \\ \quad{\text { means that } \int_{0}^{\infty} e^{-x^{2}} d x \text { can be replaced by } \int_{0}^{3} e^{-x^{2}} d x \text { without }} \\\\\quad {\text { introducing an error of magnitude greater than } 0.000042 \text { . }} \\ {\text { b. Evaluate } \int_{0}^{3} e^{-x^{2}} d x \text { numerically. }}\end{array} \end{equation}

Use reduction formulas to evaluate the integrals in Exercises \(41-50 .\) $$ \int \csc ^{5} x d x $$

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