Question

use integration by parts to derive the given formula. $$ \int x^{\alpha} \ln x d x=\frac{x^{\alpha+1}}{\alpha+1} \ln x-\frac{x^{\alpha+1}}{(\alpha+1)^{2}}+C, \alpha \neq-1 $$

Short answer

The derived formula is confirmed as correct.

Step-by-step solution

Identify Parts

The formula for integration by parts is \( \int u \, dv = uv - \int v \, du \). First, we need to choose \( u \) and \( dv \). Here, let \( u = \ln x \) and \( dv = x^{\alpha} \, dx \).

Differentiate and Integrate

Differentiate \( u \) to get \( du = \frac{1}{x} \, dx \). Integrate \( dv \) to get \( v = \frac{x^{\alpha+1}}{\alpha+1} \), where \( \alpha eq -1 \).

Apply Integration by Parts

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

Simplify the Expression

The expression can be rewritten as: \[ \frac{x^{\alpha+1}}{\alpha+1} \ln x - \frac{1}{\alpha+1} \int x^{\alpha} \, dx \]

Integrate the Remaining Part

Integrate \( \int x^{\alpha} \, dx \), obtaining \( \frac{x^{\alpha+1}}{\alpha+1} \). The expression becomes: \[ \frac{x^{\alpha+1}}{\alpha+1} \ln x - \frac{1}{\alpha+1} \cdot \frac{x^{\alpha+1}}{\alpha+1} + C \]

Final Simplification

Simplify the expression to obtain the result: \[ \frac{x^{\alpha+1}}{\alpha+1} \ln x - \frac{x^{\alpha+1}}{(\alpha+1)^2} + C \]

Key concepts

Calculus

Calculus is a branch of mathematics that focuses on change. It involves exploring ideas related to rates of change and accumulation. These concepts are crucial in understanding various phenomena in physics, engineering, finance, and many other fields.

• Differential Calculus: This analyzes how functions change. By using derivatives, it finds rates of change, which are crucial for understanding slopes and motion.
• Integral Calculus: This explores the accumulation of quantities. It looks into areas under curves and is used for solving problems related to total change and space.

Calculus forms the foundation for advanced topics and applications in the real world, making it an essential tool for many disciplines.

Integral Calculus

Integral calculus is the study of integrals and their properties. It's the inverse process of differentiation, focusing on the accumulation of quantities.

• Integration: The core operation in integral calculus. It involves finding a function whose derivative matches a given function. It helps calculate areas under curves, volumes, and other accumulative measurements.
• Definite and Indefinite Integrals: An indefinite integral involves finding an antiderivative and includes a constant of integration \( C \). A definite integral computes the accumulation between two limits, providing a specific numerical value.

A helpful technique in integration is integration by parts, based on the product rule of differentiation. It's useful for integrating products of functions by transforming them into simpler integrals. This technique is particularly handy when dealing with functions like \( x^\alpha \ln x \), as seen in the original exercise.

Mathematical Proofs

Mathematical proofs provide the foundation for validating mathematical statements. They ensure that conclusions are logically derived from assumptions and previously proven theorems.

• Structure of Proofs: A typical proof starts with given information, follows with logical reasoning using known principles, and concludes with a derived result. This process confirms the statement’s validity.
• Importance in Calculus: Proofs in calculus, like those involving integration by parts, establish the veracity of formulas and methods. They show why particular approaches work, ensuring deeper comprehension and reliability.

In the context of the original exercise, the step-by-step solution acted as a form of proof. It demonstrated the reliability of the integration by parts method for deriving the given formula, validating the approach and outcome.