Question

Find the antiderivative. \(\int x \ln \left(x^{2}\right) d x\)

Short answer

The antiderivative is \( \ln(x) x^2 - \frac{x^2}{2} + C \).

Step-by-step solution

Identify the Integration Method

The integrand is of the form \( x \ln\left(x^2\right) \). We can simplify the expression using the log property: \( \ln(a^b) = b\ln(a) \). So, \( \ln\left(x^2\right) = 2\ln(x) \). This means we need to integrate \( \int x \cdot 2\ln(x) \, dx \). Recognizing that this involves a product of a polynomial and a logarithmic function, we choose to integrate by parts.

Apply Integration by Parts Formula

The formula for integration by parts is \( \int u \, dv = uv - \int v \, du \). Here, we can choose \( u = \ln(x) \) and \( dv = 2x \, dx \). This means \( du = \frac{1}{x} \, dx \) and \( v = x^2 \), since the integral of \( 2x \, dx \) is \( x^2 \).

Substitute into Integration by Parts

Substitute the values into the integration by parts formula: \[\int x \cdot 2\ln(x) \, dx = \ln(x) \cdot x^2 - \int (x^2 \cdot \frac{1}{x}) \, dx \]Simplifying the remaining integral gives us:\[\ln(x) \cdot x^2 - \int x \, dx\]

Integrate the Remaining Simple Integral

Now solve the integral \( \int x \, dx \), which is a basic integral: \[\int x \, dx = \frac{x^2}{2} \]

Write the Final Antiderivative

Substitute back the result from Step 4 into the expression in Step 3 to find the antiderivative:\[\ln(x) \cdot x^2 - \frac{x^2}{2} + C\]This is the complete antiderivative of the original integral. Here, \( C \) is the constant of integration, needed for indefinite integrals.

Key concepts

Antiderivative

The antiderivative, also known as the indefinite integral, is a fundamental concept in calculus. It is the reverse process of differentiation. When you find the antiderivative, you are looking for a function whose derivative matches a given function. This helps in understanding the area under a curve or solving differential equations.
Finding the antiderivative can involve various techniques depending on the type of function involved, such as polynomials, trigonometric functions, or logarithms. For complex expressions, like the product of different types of functions, specific methods like integration by parts become handy.

• When you integrate, remember to add the constant of integration, denoted by \( C \), because differentiating will remove any constant present.
• Antidifferentiation requires careful consideration of function types and the appropriate techniques to use.

Logarithmic Function

The logarithmic function, denoted as \( \ln(x) \), represents the inverse operation of exponentiation. In simpler terms, it answers the question: "To what power must a base number be raised to get another number?" Understanding the properties of logarithms, like the power rule \( \ln(a^b) = b\ln(a) \), is crucial for integrating functions involving logarithms.
Logarithmic functions are prevalent in calculating growth and decay, solving exponential equations, and are encountered in many calculus problems. When combined with polynomials, they form integrals that are perfectly suited for techniques like integration by parts.

• Remember that \( \ln(x) \) is defined only for \( x > 0 \).
• Simplifying expressions with logarithmic properties can make integration much easier.

Polynomial Integration

Polynomial integration involves finding the antiderivative of polynomial expressions, which is typically straightforward. Each term of a polynomial is integrated by increasing the exponent by one and dividing by the new exponent. For example, the integral of \( x^n \) is \( \frac{x^{n+1}}{n+1} \), provided \( n \) is not equal to -1.
When polynomials interact with other functions, such as logarithms or trigonometric functions, techniques like integration by parts or substitution come into play. These methods help resolve integrals that are not direct polynomial expressions yet contain polynomial terms.

• Always check for basic integration rules to simplify terms quickly.
• Consider polynomial degree reductions and factor simplifications to assist in integration.

Calculus Problem Solving

Calculus problem solving often requires a blend of different techniques and a good understanding of when and how to apply them. In the given example, integration by parts was necessary because of the interplay between the polynomial \( x \) and the logarithmic function \( \ln(x) \).
The key to tackling such problems is recognizing patterns in the integrand and knowing the relevant formulas. Methods like integration by parts, substitution, or even transformations are tools to simplify and solve complex integrals.

• Identify if multiple techniques can be combined for a single integral.
• Practice makes perfect; regularly solving problems enhances intuition for the best method to use.

Developing a toolkit of strategies allows you to approach any calculus problem with confidence, ensuring that integrations, especially those involving products of diverse function types, are solvable with some effort and insight.