Question

derive the given reduction formula using integration by parts. $$ \int(\ln x)^{\alpha} d x=x(\ln x)^{\alpha}-\alpha \int(\ln x)^{\alpha-1} d x $$

Short answer

The reduction formula is derived using integration by parts: \( u = (\ln x)^{\alpha} \), \( dv = dx \).

Step-by-step solution

Understand the Formula

The problem is to derive the reduction formula using integration by parts: \( \int (\ln x)^{\alpha} \, dx = x(\ln x)^{\alpha} - \alpha \int (\ln x)^{\alpha-1} \, dx \). This formula helps reduce the power of \((\ln x)\) in integrals.

Set Up Integration by Parts

Recall the integration by parts formula: \( \int u \, dv = uv - \int v \, du \). Choose \( u = (\ln x)^{\alpha} \) and \( dv = dx \). This choice helps simplify the integral.

Differentiate and Integrate Parts

Calculate \( du \) and \( v \):- Differentiate \( u = (\ln x)^{\alpha} \): \( du = \alpha (\ln x)^{\alpha-1} \cdot \frac{1}{x} \, dx \) using the chain rule.- Integrate \( dv = dx \): \( v = x \).

Apply Integration by Parts

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

Simplify the Expression

The second term of the integral simplifies to \( \alpha \int (\ln x)^{\alpha-1} \, dx \), because \( x \cdot \frac{1}{x} = 1 \). Thus, the integration by parts derives the reduction formula: \( \int (\ln x)^{\alpha} \, dx = x(\ln x)^{\alpha} - \alpha \int (\ln x)^{\alpha-1} \, dx \).

Key concepts

Reduction Formula

A reduction formula is a powerful tool in integral calculus that simplifies the process of integrating functions raised to a power. It creates a recursive relationship, helping to reduce complex integrals into simpler ones that can be evaluated step-by-step. In the context of the given exercise, the reduction formula helps us manage integrals like \( \int(\ln x)^\alpha \, dx \) by reducing the exponent \( \alpha \) incrementally in each step.

This method offers a strategic approach to tackling powers and logarithms, where otherwise, finding an antiderivative might not be straightforward. By using the reduction formula, you are able to express the integral of \( (\ln x)^\alpha \) in terms of \( (\ln x)^{\alpha-1} \), thus providing a simpler calculation pathway.

• The formula takes the form: \( \int(\ln x)^{\alpha} \, dx = x(\ln x)^{\alpha} - \alpha \int(\ln x)^{\alpha-1} \, dx \).
• This recursive nature is key to solving these types of integrals efficiently.

Chain Rule

The chain rule is a fundamental principle in calculus used to differentiate compositions of functions. It provides a method for calculating the derivative of a composite function and is essential for differentiating the chosen function \( u = (\ln x)^\alpha \) in the integration by parts process.

When you're dealing with \( (\ln x)^{\alpha} \), the chain rule allows us to handle the power of \( \ln x \) by differentiating it correctly. You start by taking the derivative of the outer function, which is \( (\ln x)^{\alpha} \), with respect to the inner function, and then multiply by the derivative of the inner function \( \ln x \).

• Differentiate: \( u = (\ln x)^\alpha \).
• Using the chain rule, we find: \( du = \alpha(\ln x)^{\alpha-1} \cdot \frac{1}{x} \, dx \).
• This step prepares us for integrating \( v = x \) in the integration by parts formula.

Understanding and applying the chain rule correctly ensures the term \( du \) is properly calculated, setting the stage for successful integration by parts.

Integral Calculus

Integral calculus is an entire branch of mathematics dedicated to finding antiderivatives and integrals, offering solutions to a plethora of real-world applications. In this exercise, our focus is on evaluating definite and indefinite integrals, particularly those involving complex expressions like logarithms raised to a power.

The power of integral calculus lies in its ability to find solutions to growth, decay, area, and motion, among others. By using techniques such as integration by parts, as shown in this reduction formula exercise, we are able to break down complex integrals into manageable parts.

• Integral calculus connects seemingly disparate parts of an equation, summing up infinitesimal components.
• Through methods like substitution and integration by parts, we unlock insights into areas under curves and accumulated quantities.
• This exercise showcases the essential role of integration techniques in simplifying and evaluating more complex expressions.

Integral calculus is not just about solving equations; it’s about understanding the deeper relationships between integrated components within functions.

Logarithmic Integration

Logarithmic integration deals with integrating expressions that involve logarithmic functions, like \( \ln x \), which often appear in various mathematical contexts. They provide challenges due to their non-polynomial nature, and integrating such functions often requires special techniques or substitutions.

In our problem, the logarithmic function \( (\ln x)^\alpha \) needs special attention within the integral setup, typically seen in situations where conventional integration techniques would not straightforwardly apply. Integrating such functions can be complex, yet integration by parts offers a pathway to handle these expressions effectively.

• Logarithms add complexity to integration because they do not simplify easily through basic algebraic manipulation.
• In our exercise, we handle \( \ln x \) by assigning it a pivotal role in the function \( u \) for integration by parts: \( u = (\ln x)^\alpha \).
• This strategic choice helps differentiate and then integrate the function efficiently, leading to the derived reduction formula.

This approach highlights the importance of strategic method applications when integrating logarithmic functions, a core aspect of mastering integral calculus.