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Chapter 8: Problem 66

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 is verified using integration by parts.

Step by step solution

01

Identify the functions for integration by parts

The integration by parts formula is \( \int u \, dv = uv - \int v \, du \). Here, we will let \( u = (\ln x)^n \), making \( du = n(\ln x)^{n-1} \frac{1}{x} \, dx \), and \( dv = dx \), making \( v = x \).
02

Apply the integration by parts formula

Substitute \( u \), \( du \), \( v \), and \( dv \) into the integration by parts formula:\[\int (\ln x)^n \, dx = x(\ln x)^n - \int x \cdot n(\ln x)^{n-1} \frac{1}{x} \, dx\]Simplifying inside the integral, this becomes:\[\int (\ln x)^n \, dx = x(\ln x)^n - n \int (\ln x)^{n-1} \, dx\]
03

Verify the reduction formula

Both sides of the equation derived:\[x(\ln x)^n - n \int (\ln x)^{n-1} \, dx\]Matches the given reduction formula:\[\int(\ln x)^{n} \, d x=x(\ln x)^{n}-n \int(\ln x)^{n-1} \, d x\]This confirms that our integration by parts application correctly established the reduction formula.

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Key Concepts

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

Integration by Parts
The technique of integration by parts is a powerful tool in calculus, especially useful when dealing with the product of functions. The formula \[ \int u \, dv = uv - \int v \, du \]is a way to transform a complex integral into simpler parts.
Here's how the process works:
  • Select a part of the integrand as \( u \) (a part to differentiate) and the remaining as \( dv \) (a part to integrate).
  • Differentiate \( u \) to find \( du \), and integrate \( dv \) to find \( v \).
  • Replace these components into the formula to form a new integral, usually simpler than the original.
Integration by parts is often applied multiple times and can lead to the development of reduction formulas, making it a cornerstone technique in advanced calculus.
Reduction Formulas in Integration
Reduction formulas are algebraic identities that simplify the integration of power functions, often by relating the integral of a higher power to a lower one. These formulas provide a recursive method for tackling otherwise intractable integrals.
  • They help reduce complex integrals into a chain of simpler ones, making calculations more manageable.
  • Reduction formulas usually emerge naturally from repeated application of techniques like integration by parts.
  • In the case of the exercise provided, it simplifies integrating powers of the logarithmic function \((\ln x)^n\).
This makes reduction formulas an important concept not only for manual integration techniques but also for computing solutions efficiently in advanced topics.
Logarithmic Integration
Logarithmic integration involves integrating expressions containing \( \ln x \). These integrals can often seem intimidating owing to the logarithm's properties but can be simplified with proper techniques.
  • Integration by parts is particularly useful here as logarithms typically simplify when paired with polynomial functions.
  • Choosing the logarithmic term as \( u \) in integration by parts often helps reduce the complexity of the integral.
  • Careful application and manipulation with this tool can also lead to the discovery and verification of reduction formulas, as seen in our exercise scenario.
Mastering logarithmic integration is crucial in calculus as it lays the foundation for handling more complex transcendental functions.
Calculus Problems
In calculus problems, especially those involving integrals, understanding core techniques is vital. These problems often build upon foundational concepts to develop deeper problem-solving skills.
  • Each problem type aids in understanding the underlying principles of calculus when approached methodically.
  • Identifying the right technique, such as integration by parts, can greatly simplify the solution.
  • Proficiency in solving these problems enhances one's ability to tackle real-world applications, including those involving logarithmic and transcendental functions.
Deepening the understanding of these calculus problems ensures you're well-equipped for both academic and practical challenges involving integration and beyond.

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Most popular questions from this chapter

In Exercises \(9-16,\) express the integrand as a sum of partial fractions and evaluate the integrals. $$\int_{4}^{8} \frac{y d y}{y^{2}-2 y-3}$$

In Exercises \(21-32,\) express the integrand as a sum of partial fractions and evaluate the integrals. $$\int \frac{2 \theta^{3}+5 \theta^{2}+8 \theta+4}{\left(\theta^{2}+2 \theta+2\right)^{2}} d \theta$$

In Exercises \(9-16,\) express the integrand as a sum of partial fractions and evaluate the integrals. $$\int \frac{2 x+1}{x^{2}-7 x+12} d x$$

As Example 3 shows, the integral \(\int_{1}^{\infty}(d x / x)\) diverges. This means that the integral $$\int_{1}^{\infty} 2 \pi \frac{1}{x} \sqrt{1+\frac{1}{x^{4}}} d x$$ which measures the surface area of the solid of revolution traced out by revolving the curve \(y=1 / x, 1 \leq x,\) about the \(x\)-axis, diverges also. By comparing the two integrals, we see that, for every finite value \(b>1,\) $$\int_{1}^{b} 2 \pi \frac{1}{x} \sqrt{1+\frac{1}{x^{4}}} d x>2 \pi \int_{1}^{b} \frac{1}{x} d x$$ (Check your book to see image) However, the integral $$\int_{1}^{\infty} \pi\left(\frac{1}{x}\right)^{2} d x$$ for the volume of the solid converges. a. Calculate it. b. This solid of revolution is sometimes described as a can that does not hold enough paint to cover its own interior. Think about that for a moment. It is common sense that a finite amount of paint cannot cover an infinite surface. But if we fill the horn with paint (a finite amount), then we will have covered an infinite surface. Explain the apparent contradiction.

In Exercises \(17-20\), express the integrand as a sum of partial fractions and evaluate the integrals. $$\int \frac{d x}{\left(x^{2}-1\right)^{2}}$$
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