Question

State whether you would use integration by parts to evaluate the integral. If so, identify u and dv. If not, describe the technique used to perform the integration without actually doing the problem. $$ \int \frac{\ln ^{2} x}{x} d x $$

Short answer

Use integration by parts with \( u = \ln^2 x \) and \( dv = \frac{1}{x} \, dx \).

Step-by-step solution

Identify the Type of Integral

The given integral is \( \int \frac{\ln^2 x}{x} \, dx \). Notice that it contains \( \ln x \), which typically suggests considering integration by parts.

Review the Integration by Parts Formula

The integration by parts formula is \( \int u \, dv = uv - \int v \, du \). To apply this, we need to identify suitable \( u \) and \( dv \) such that the differentiation and new integral become simpler.

Select \( u \) and \( dv \)

In the integral \( \int \frac{\ln^2 x}{x} \, dx \), set \( u = \ln^2 x \), because differentiating \( u \) reduces the power of the logarithmic term. Then set \( dv = \frac{1}{x} \, dx \), which is straightforward to integrate.

Verify the Choice of \( u \) and \( dv \)

Differentiate \( u \) to get \( du = 2 \ln x \cdot \frac{1}{x} \, dx \), and integrate \( dv \) to get \( v = \ln x \). Both operations yield simpler expressions, confirming this as a good choice.

Key concepts

Integration Techniques

When approaching integrals, several techniques can be used to simplify and solve them. Sometimes, basic integration methods aren't enough, especially when faced with complex functions. This is where advanced integration techniques like substitution, partial fraction decomposition, and integration by parts come into play.

Let's focus on integration by parts, which is especially useful when dealing with products of functions, such as a polynomial times a logarithmic function. The integration by parts formula is:

• \( \int u \, dv = uv - \int v \, du \)

The goal is to pick parts of the integral to be \( u \) and \( dv \) so that the resulting \( \int v \, du \) is simpler than the original. A helpful mnemonic for choosing \( u \) and \( dv \) is "LIATE", which stands for Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, and Exponential functions. Typically, the order from left to right suggests which function to choose as \( u \).

In the integration of \( \int \frac{\ln^2 x}{x} \, dx \), the presence of \( \ln x \) makes integration by parts a viable option, as isolating the logarithmic part as \( u \) often simplifies the problem. Thus, selecting \( u = \ln^2 x \) and \( dv = \frac{1}{x} \, dx \) is appropriate.

Logarithmic Functions

Logarithmic functions, particularly the natural logarithm \( \ln x \), are prevalent in calculus. They are often involved in mathematical models relating to growth processes and decay phenomena due to their properties.

When dealing with logarithmic functions in integration, applying the right technique is crucial. Since the derivative of \( \ln x \) is \( \frac{1}{x} \), these functions tend to simplify easily, either lending themselves to straightforward substitution or to integration by parts.

In the given integral \( \int \frac{\ln^2 x}{x} \, dx \), the square of the logarithmic function suggests compounded complexity. Here, the natural choice is to apply integration by parts, picking \( \ln^2 x \) as \( u \), because differentiation reduces the power, hence simplifying the expression. This illustrates how understanding the behavior of logarithmic functions is key to applying the most effective integration technique.

Definite Integrals

Definite integrals differ from indefinite integrals by providing a specific value, not just a function family. They represent the accumulation of quantities, often used to calculate areas under curves or between functions.

The process of evaluating a definite integral involves computing the antiderivative and applying limits. Although our example \( \int \frac{\ln^2 x}{x} \, dx \) doesn't immediately consider limits, the concept remains crucial.

When faced with a definite integral, after finding the antiderivative using the appropriate technique (like integration by parts), the next step is to substitute the limits of integration. This involves finding the difference between the values of the antiderivative at the upper and lower limits. In practice, understanding the properties of definite integrals streamlines calculations for real-world applications such as finding volumes, work, and other cumulative measures. This robustness further demonstrates why mastering integral calculus is highly valuable.