Problem 66 - Calculus Early Transcendentals

Question

Miscellaneous integrals Evaluate the following integrals. \(\int \frac{\ln ^{2} x+2 \ln x-1}{x} d x\)

Step-by-Step Solution

Answer

The integral of the given function is: $$\int \frac{\ln ^{2} x+2 \ln x-1}{x} d x = \left(\frac{1}{3}\ln^3 x + (\ln x)^2 -\ln|x|\right) + C$$, where \(C\) is the constant of integration.

Step 1: Rewrite the integrand using logarithmic properties

The given integral is: $$\int \frac{\ln ^{2} x+2 \ln x-1}{x} d x$$ Let's rewrite the integral by separating the terms and considering the properties of logarithms: $$\int \left(\frac{\ln ^{2} x}{x} +\frac{2 \ln x}{x} -\frac{1}{x}\right) d x$$ Now we can deal with each term separately.

Step 2: Substitution for the first term

Let's find the integral of the first term: $$\int \frac{\ln ^{2} x}{x} d x$$ To evaluate this integral, we can use the substitution: $$u = \ln x \Rightarrow \frac{d u}{d x} = \frac{1}{x} \Rightarrow d u = \frac{1}{x} d x$$ With this substitution, the integral becomes: $$\int u^2 d u$$ Now, we can easily integrate: $$\int u^2 d u = \frac{1}{3}u^3 + C_1 = \frac{1}{3}\ln^3 x + C_1$$

Step 3: Integrate the second term

The second term can be easily integrated using the logarithmic property that states \((\ln x)^\prime = 1/x\). $$\int \frac{2\ln x}{x} dx = 2\int \ln x \left(\frac{1}{x}dx\right) = 2\int \ln x d(\ln x) = 2\frac{1}{2}(\ln x)^2 + C_2 = (\ln x)^2 + C_2$$

Step 4: Integrate the third term

The third term is the integral of \(-\frac{1}{x}\), which can be easily integrated as well: $$\int -\frac{1}{x} d x = -\int \frac{1}{x} d x = -\ln|x| + C_3$$

Step 5: Sum up the results

Now we can sum up the results of each term to find the final solution: $$\int \frac{\ln ^{2} x+2 \ln x-1}{x} d x = \left(\frac{1}{3}\ln^3 x + (\ln x)^2 -\ln|x|\right) + C$$ Where \(C = C_1 + C_2 + C_3\) is the constant of integration.

Key Concepts

Logarithmic Integration

Logarithmic integration is useful when dealing with integrals involving logarithmic functions. In our exercise, we have an expression \[ \int \frac{\ln ^{2} x+2 \ln x-1}{x} \; dx \] which appears complex at first but can be simplified by breaking down its terms. By rewriting the integrand, we apply properties of logarithms to separate the expression into simpler parts:- \( \int \frac{\ln^2 x}{x} \; dx \)- \( \int \frac{2 \ln x}{x} \; dx \)- \( \int -\frac{1}{x} dx \)Each term can then be addressed with different techniques, such as substitution and basic integration rules. Understanding how to manipulate logarithms and recognizing these forms is crucial for mastering integration problems involving logs.

Substitution Method

The substitution method is a powerful technique in calculus that simplifies complex integrals by changing variables. In our problem, for the term \( \int \frac{\ln^2 x}{x} \; dx \), we can set \( u = \ln x \), which gives us \( du = \frac{1}{x} dx \). This substitution transforms the integral into \( \int u^2 du \), which is much easier to solve. Substitution helps transform the problem into a simpler form, making it easier to integrate. By choosing the right substitution, you can often reduce the complexity of the original integral. Remember, the key is to express both the integrand and the differential in terms of the new variable to solve the integral smoothly. Once integrated, always remember to revert back to the original variable to provide the solution in the initial context.

Calculus

Calculus, the mathematical study of change, provides robust tools for solving integrals. Integration is one of the core processes in calculus. It allows us to find the area under curves, interpret differential equations, and assess continuous growth. In our exercise, we have dissected a complex logarithmic integral into simpler parts using calculus principles:

• Separation of terms and handling them individually using basic integration rules and substitutions.
• Applying knowledge of derivatives, recognizing that the derivative of \( \ln x \) is \( \frac{1}{x} \).
• Summing resulting integrals and incorporating constant of integration, \( C \).

These steps illustrate how calculus is not only about computing values but understanding functions' behaviors and transformations. Each technique builds on foundational calculus concepts, underscoring their importance in solving real-world mathematical problems.