Problem 42 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The reduction formula is established as \(\int(\ln x)^{n} \, dx=x(\ln x)^{n}-n \int(\ln x)^{n-1} \, dx\).

Step 1: Integration by Parts Formula

The integration by parts formula is given by \[\int u \, dv = uv - \int v \, du.\] We will use this formula to establish the reduction formula for \(\int(\ln x)^{n} \, dx.\).

Step 2: Choose u and dv

Set \( u = (\ln x)^n \) and \( dv = dx \). This implies \( du = n(\ln x)^{n-1} \frac{1}{x} dx \) and \( v = x \).

Step 3: Apply Integration by Parts

Substitute \( u, \ du, \ v, \ dv \) into the integration by parts formula: \[ \int (\ln x)^n \, dx = x(\ln x)^n - \int x \cdot n(\ln x)^{n-1} \frac{1}{x} \, dx. \] Simplify the resulting integrand.

Step 4: Simplify the Integral

The integral becomes \[ \int (\ln x)^n \, dx = x(\ln x)^n - n \int (\ln x)^{n-1} \, dx. \] Thus, we have established the reduction formula.

Key Concepts

Reduction Formula

A reduction formula is a pattern or recursive relation in integration that simplifies the process of evaluating integrals by reducing them into simpler forms. This technique is especially useful for complex integrals that would otherwise be very complicated to solve directly. The key idea behind using a reduction formula is to express an integral of a higher complexity in terms of a similar integral that is less complex.

In our example, the reduction formula deals with \[\int(\ln x)^{n} \, dx = x(\ln x)^{n} - n \int(\ln x)^{n-1} \, dx.\]This means we can express the integral of \((\ln x)^{n}\)in terms of the integral of \((\ln x)^{n-1}.\)By doing this, each successive application of the formula decreases the power of \(n\) until we reach an integral that can be evaluated directly.

Using a reduction formula is not only a valuable tool in solving polynomial integrals but also assists in approaching complex integrals in a structured way. It simplifies the work, ensuring that each stage becomes more manageable.

Integration Techniques

Integration techniques encompass various methods used to find integrals, each tailored to different forms of functions. The Integration by Parts method is one of these essential techniques. It is based on the product rule for differentiation and is particularly useful when dealing with products of functions that are difficult to integrate directly.

• **Substitution Method:** Changes variables to simplify the integral.
• **Partial Fraction Decomposition:** Breaks down complex fractions into simpler parts.
• **Trigonometric Substitution:** Simplifies integrals involving square roots using trigonometric identities.

Using integration by parts usually involves choosing which function to differentiate and which to integrate. For \(\int u \, dv = uv - \int v \, du\), identifying \(u\) and \(dv\) strategically is key. Consider the functions involved in the integral to determine which choice simplifies the problem more effectively. In our exercise, selecting \(u = (\ln x)^n\)\ and \(dv = dx\)\ was critical to establish the reduction formula.

Natural Logarithm

The natural logarithm, denoted as \(\ln x\), is a logarithmic function with the base \(e\), where \(e\) is approximately equal to 2.71828. It plays a significant role in calculus, particularly due to its derivative and integration properties.

• **Derivative of \(\ln x\):** The derivative of \(\ln x\) is \(1/x\). This simple derivative makes it easier to differentiate functions involving \(\ln x\).
• **Integral of \(\ln x\):** The integral of \(\ln x\) alone is not straightforward and requires the use of integration by parts.

In the realm of advanced mathematics, the natural logarithm appears frequently in solutions involving exponential growth and decay, as well as in processes involving compounding continuously. Understanding how to manipulate expressions with \(\ln x\)\ is vital in higher-level calculus as it aids in simplifying complex expressions and integrating challenging functions. Indeed, in our specific example, \(\ln x\)\ being raised to a power necessitates using techniques like integration by parts to handle the intricacies involved in integrating such terms.