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Use integration by parts to establish the reduction formula. $$\int(\ln x)^{n} d x=x(\ln x)^{n}-n \int(\ln x)^{n-1} d x$$

Short Answer

Expert verified
The reduction formula using integration by parts is correctly established as \(\int (\ln x)^n dx = x(\ln x)^n - n \int (\ln x)^{n-1} dx\).

Step by step solution

01

Identify Integration by Parts Formula

Recall the integration by parts formula: \[\int u\, dv = uv - \int v\, du\]. Choose suitable \(u\) and \(dv\) such that \(du\) and \(v\) are easy to compute.
02

Choose Components for u and dv

Let \(u = (\ln x)^n\) and \(dv = dx\). Then, differentiate \(u\) and integrate \(dv\):\[du = n(\ln x)^{n-1} \cdot \frac{1}{x} dx\] and \[v = x.\]
03

Apply Integration by Parts

Use the integration by parts formula with the chosen \(u\) and \(v\): \[\int (\ln x)^n dx = x(\ln x)^n - \int x \cdot n(\ln x)^{n-1} \cdot \frac{1}{x} dx.\] Simplify the integral: \[= x(\ln x)^n - n \int (\ln x)^{n-1} dx.\]
04

Establish the Reduction Formula

Observe that the left side matches the original integral we are looking for: \[\int (\ln x)^n dx = x(\ln x)^n - n \int (\ln x)^{n-1} dx.\] Thus, the reduction formula is established as shown in the exercise prompt.

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Key Concepts

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

Reduction Formula
A reduction formula in integrals simplifies the evaluation of integrals that might be complex to integrate directly. Using integration by parts helps in formulating this type of equation.

The key idea is that by expressing an integral in terms of a simpler one, the task of integration becomes easier. For instance, when dealing with powers of logarithmic functions, a reduction formula allows us to tackle integrals progressively. A problem like finding \(\int(\ln x)^{n} dx\) is reduced to solving \(\int(\ln x)^{n-1} dx\), which is a simpler version.

This reduction approach is extremely helpful in calculus where repetitive application of integration techniques can simplify a problem into a manageable form. Remember, identifying patterns in integration problems is key to applying a reduction formula.
  • It's useful for problems involving powers or repetitive elements.
  • Aids in systematically breaking down complex expressions.
Calculus
Calculus is the mathematical study of continuous change, and its two main branches are differential and integral calculus. This problem focuses on integral calculus, specifically integration by parts.

Integral calculus is widely used to determine areas, volumes, and solutions to differential equations. Understanding which technique to apply is crucial. In this exercise, integration by parts was chosen due to the involvement of a product of functions, specifically \(u = (\ln x)^n\) and \(dv = dx\).

When solving integrals:
  • Identify if product or composite functions exist.
  • Choose \(u\) and \(dv\) correctly to ensure that \(du\) and \(v\) remain simple.
  • Simplification might involve breaking down the left-side integral to match the original structure.
You’ll often find such methods very effective in problems involving natural logarithms or exponential functions.
Logarithmic Integration
Logarithmic integration involves integrating functions that contain logarithmic expressions. As logs can be trickier than polynomial expressions, strategies like integration by parts become essential to solve these effectively.

The main challenge is managing expressions such as \((\ln x)^n\), where \(n\) is a power making straightforward integration difficult. Instead, breaking down the expression or using reductions makes the integration process feasible.

Key reminders for logarithmic integration include:
  • Logarithms require different rules than polynomials or trigonometric functions.
  • Careful selection of \(u\) and \(dv\) aids in simplification especially when dealing with \(\ln x\).
Mastering these techniques enhances your ability to handle sophisticated mathematical problems, including those outside direct calculus applications.

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Most popular questions from this chapter

Use integration, the Direct Comparison Test, or the limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer. $$\int_{0}^{1} \frac{\ln x}{x^{2}} d x$$

A taxicab company in New York City analyzed the daily number of kilometers driven by each of its drivers. It found the average distance was \(300 \mathrm{km}\) with a standard deviation of \(50 \mathrm{km}\). Assuming a normal distribution, what prediction can we make about the percentage of drivers who will log in either more than \(400 \mathrm{km}\) or less than \(250 \mathrm{km} ?\)

Use a CAS to perform the integrations. Evaluate the integrals $$\text { a. } \int \frac{\ln x}{x^{2}} d x \quad \text { b. } \int \frac{\ln x}{x^{3}} d x \quad \text { c. } \int \frac{\ln x}{x^{4}} d x$$ d. What pattern do you see? Predict the formula for $$\int \frac{\ln x}{x^{5}} d x$$ and then see if you are correct by evaluating it with a CAS. e. What is the formula for $$\int \frac{\ln x}{x^{n}} d x, \quad n \geq 2 ?$$ Check your answer using a CAS.

Compare the results of using Equations (4) and (5). Equations (4) and (5) lead to different formulas for the integral of \(\tan ^{-1} x\) a. \(\int \tan ^{-1} x d x=x \tan ^{-1} x-\ln \sec \left(\tan ^{-1} x\right)+C\) Eq. (4) b. \(\int \tan ^{-1} x d x=x \tan ^{-1} x-\ln \sqrt{1+x^{2}}+C\) Eq. (5) Can both integrations be correct? Explain.

Use integration, the Direct Comparison Test, or the limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer. $$\int_{1}^{\infty} \frac{1}{e^{x}-2^{x}} d x$$

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