Problem 65 - Calculus Single Variable

Question

Calculate the integrals. $$ \int\left(\frac{2}{x \ln \left(x^{2}\right)}-\frac{2}{x \ln ^{2}(x)}\right) d x $$

Step-by-Step Solution

Answer

\( \ln|\ln(x)| - \frac{2}{\ln(x)} + C \)

Step 1: Simplify the Integrand

First, simplify each term in the integrand. Start by examining the first term: \(\frac{2}{x \ln(x^2)} = \frac{2}{x \cdot 2 \ln(x)} = \frac{1}{x \ln(x)}\). Now the expression inside the integrand becomes \(\frac{1}{x \ln(x)} - \frac{2}{x \ln^2(x)}\).

Step 2: Integrate Each Term Separately

The integrand can be split into two separate integrals: \[ \int \frac{1}{x \ln(x)} \, dx - \int \frac{2}{x \ln^2(x)} \, dx. \]Integrate these individually.

Step 3: Perform Substitution for First Integral

For \( \int \frac{1}{x \ln(x)} \, dx \), we use substitution. Let \( u = \ln(x) \), hence \( du = \frac{1}{x} \, dx \). The integral becomes:\[ \int \frac{1}{u} \, du = \ln|u| + C_1 = \ln|\ln(x)| + C_1. \]

Step 4: Perform Substitution for Second Integral

For \( \int \frac{2}{x \ln^2(x)} \, dx \), use the same substitution \( u = \ln(x) \), and \( du = \frac{1}{x} \, dx \). The integral becomes:\[ \int \frac{2}{u^2} \, du = -\frac{2}{u} + C_2 = -\frac{2}{\ln(x)} + C_2. \]

Step 5: Combine Results

Combine the results of the integrals from Steps 3 and 4:\[ \ln|\ln(x)| - \frac{2}{\ln(x)} + C, \]where \(C = C_1 + C_2\) is the constant of integration.

Key Concepts

Integration Techniques

Integral calculus is a cornerstone of mathematical study and involves methods for finding the area under a curve. When faced with a complex integrand, different integration techniques can simplify the process. One such method is breaking down the integrand into simpler parts that are easier to manage. This is often done by identifying terms that can be separated and integrated individually.

This approach is used particularly when the integration can be split into a sum or difference of simpler parts. For example, you can tackle each term of the integrand separately, as seen in our example exercise where the integrand is divided into two fractions.

When choosing an integration technique, consider:

• Breaking down the integrand into simpler expressions.
• Identifying terms that can be integrated separately.
• Using algebraic manipulations to simplify the integrand.

These techniques not only simplify the computational process but also deepen your understanding of function behaviors.

Substitution Method

The substitution method is a powerful tool in integration, often likened to the chain rule in differential calculus. It involves replacing a part of the integrand with a variable to simplify the function that needs to be integrated.

For example, in the first term of our exercise, the substitution was: let \( u = \ln(x) \), and therefore \( du = \frac{1}{x} \, dx \). This change of variables transforms the integral into one involving the variable \( u \), which is simpler to evaluate.

A step-by-step approach to substitution:

• Identify a part of the integrand that complicates the integration.
• Replace it with a substitution that simplifies the integrand.
• Differentiate the substitution to obtain \( du \).
• Rewrite the integral in terms of \( u \).
• Integrate with respect to \( u \) and substitute back to the original variable.

This technique is especially useful for integrals involving logarithms or trigonometric functions.

Definite Integrals

While the exercise at hand focuses on indefinite integrals, understanding definite integrals is just as crucial. Definite integrals determine the net area under a curve between two points on the x-axis. Unlike indefinite integrals, they yield a specific numerical result rather than a general formula with a constant of integration.

Definite integrals are often used in applications requiring precise measurements, such as finding the distance covered by an object over time.

Key points about definite integrals include:

• They have limits of integration, denoted by the upper and lower limits on the integral sign.
• The integration process involves finding the antiderivative and then evaluating it at the upper and lower limits.
• The result is \( F(b) - F(a) \), where \( F \) is the antiderivative and \( a \) and \( b \) are the limits.

Considerations around definite integrals are critical for fully utilizing integral calculus in real-world problems.

Calculus Problem Solving

Calculus problem solving often requires a combination of several techniques to reach a solution. It’s more like fitting puzzle pieces together to see the whole picture.

In our exercise, we performed simplifications and substitutions manually, skills critical for effective problem solving. An understanding of each technique's purpose and application forms a solid foundation.

Approaching calculus problems effectively involves:

• Using steps to break problems into manageable parts.
• Identifying key techniques (like substitution) that simplify the analysis.
• Recognizing patterns in the problems that allow standard methods to be applied.
• Practicing with various types of integrals to strengthen the skill set.

Successfully solving calculus problems means taking a strategic and organized approach, familiarizing oneself with different methods, and flexibly applying them.