Problem 53 - Calculus and its applications

Question

Evaluate. $$ \int x^{n}(\ln x)^{2} d x, \quad n \neq-1 $$

Step-by-Step Solution

Answer

The integral evaluates to: \( \frac{x^{n+1}}{n+1} (\ln x)^2 - \frac{2x^{n+1} \ln x}{(n+1)^2} + \frac{2x^{n+1}}{(n+1)^3} + C \).

Step 1: Identify the Type of Integration

The given integral \( \int x^n (\ln x)^2 \, dx \) involves a polynomial term \( x^n \) and a logarithmic function, which suggests the use of integration by parts.

Step 2: Set Up Integration by Parts

Recall the integration by parts formula: \( \int u \, dv = uv - \int v \, du \). Choose \( u = (\ln x)^2 \) and \( dv = x^n \, dx \).

Step 3: Compute Derivative and Determine du

Differentiate \( u = (\ln x)^2 \) using the chain rule: \( du = 2(\ln x) \cdot \frac{1}{x} \, dx = \frac{2\ln x}{x} \, dx \).

Step 4: Integrate dv to Find v

Integrate \( dv = x^n \, dx \) to find \( v \). The antiderivative is \( v = \frac{x^{n+1}}{n+1} \), assuming \( n eq -1 \).

Step 5: Substitute into Integration by Parts Formula

Substitute \( u \), \( v \), \( du \), and \( dv \) into the integration by parts formula: \[ \int x^n (\ln x)^2 \, dx = \frac{x^{n+1}}{n+1} (\ln x)^2 - \int \left( \frac{x^{n+1}}{n+1} \cdot \frac{2\ln x}{x} \right) \, dx. \]

Step 6: Simplify the Integral

Simplify the expression inside the second integral: \[ -\frac{2}{n+1} \int x^n \ln x \, dx. \] This integral again needs to be solved using integration by parts.

Step 7: Apply Integration by Parts Again

For \( \int x^n \ln x \, dx \), set \( u = \ln x \) and \( dv = x^n \, dx \). Calculate \( du = \frac{1}{x} \, dx \) and \( v = \frac{x^{n+1}}{n+1} \).

Step 8: Substitute into Integration by Parts for Second Integral

Using the integration by parts formula again, \[ \int x^n \ln x \, dx = \frac{x^{n+1}}{n+1} \ln x - \int \frac{x^{n+1}}{n+1} \cdot \frac{1}{x} \, dx. \] Simplify to get \[ \int x^n \ln x \, dx = \frac{x^{n+1}}{n+1} \ln x - \frac{1}{(n+1)^2} x^{n+1} + C. \]

Step 9: Substitute Back into Main Equation

Substitute the result from the second integral back into the original: \[ \int x^n (\ln x)^2 \, dx = \frac{x^{n+1}}{n+1} (\ln x)^2 - \frac{2}{n+1} \left( \frac{x^{n+1}}{n+1} \ln x - \frac{1}{(n+1)^2} x^{n+1} \right). \]

Step 10: Simplify and State the Final Solution

Simplify the expression by distributing and combining like terms to obtain: \[ \int x^n (\ln x)^2 \, dx = \frac{x^{n+1}}{n+1} (\ln x)^2 - \frac{2x^{n+1} \ln x}{(n+1)^2} + \frac{2x^{n+1}}{(n+1)^3} + C. \]

Key Concepts

Logarithmic Functions

Logarithmic functions, often abbreviated as logs, play a crucial role in calculus, especially when dealing with the integration of certain types of expressions. A logarithmic function is an inverse of an exponential function, commonly expressed as \( ext{ln}(x)\) in natural logarithms with base \(e\), approximately 2.718. When integrating expressions involving \( ext{ln}(x)\) or its powers, like \(( ext{ln}(x))^2\), it often requires more advanced techniques such as integration by parts. This is because directly integrating \( ext{ln}(x)\) is not straightforward, unlike basic polynomials.

• Logarithmic functions are continuous and only defined for \(x > 0\).
• The derivative of \( ext{ln}(x)\) is \(\frac{1}{x}\), a fact useful for deriving integration formulas.
• Integration involving log terms often necessitates looking for patterns or using methods such as substitution or integration by parts for feasible computation.

Understanding how logarithmic functions behave, and how they interact with other functions like polynomials, sets the stage for identifying the right integration techniques needed for solving complex integrals.

Polynomial Integration

Polynomial integration is one of the foundational skills in integral calculus. It involves finding the antiderivative of polynomial functions like \(x^n\). The process generally follows a pattern that becomes simple with practice, especially for basic or simple terms. The integral of a polynomial function \(x^n\) (where \(n eq -1\)) results in:\[\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\]where \(C\) stands for the constant of integration.

• Unlike logarithms, polynomials of the form \(x^n\) have well-defined integration rules, which make them relatively easier to handle.
• For any power of \(x\), the integral rule effectively increases the exponent by one and divides by this new exponent, summarizing a uniform process across polynomial terms.
• It's critical to understand how polynomial functions interact with other elements, as they often appear in conjunction with more complex expressions.

When polynomials are combined with other functions, like logarithmic functions, as seen in this exercise, more advanced integration methods such as integration by parts come into play, requiring attention to choosing suitable parts for integration or differentiation.

Integral Calculus

Integral calculus is a branch of calculus focusing on the accumulation of quantities and the areas under and between curves. It's an essential part of mathematical analysis and provides techniques for solving a variety of problems involving integrals and their applications.The core objective of integral calculus is to determine the antiderivative or integral of a function. For many complex functions, this involves using advanced techniques. One such technique covered in this exercise is integration by parts, based on the formula:\[\int u \, dv = uv - \int v \, du\]

• Integration by parts is especially useful for products of functions, such as log functions and polynomials, as seen in the given exercise.
• Choosing the right functions to assign as \(u\) and \(dv\) is crucial for simplifying the calculation effectively.
• This method illustrates how intricate integrals can often be deconstructed into simpler parts through systematic application of integration rules.

Integral calculus not only involves finding antiderivatives but also requires a proper understanding of how different functions interact. This includes recognizing patterns and applying the correct methods to simplify and solve integrals. Mastery of these techniques is vital for progressing through more advanced calculus problems and applications.