Question

In the following exercises, find each indefinite integral by using appropriate substitutions. $$ \int \frac{d x}{x(\ln x)^{2}} $$

Short answer

The integral is \( -\frac{1}{\ln x} + C \).

Step-by-step solution

Identify the Substitution

First, we observe that the expression inside the integral, \( \ln x \), is a good candidate for substitution because its derivative \( \frac{1}{x} \) appears in the integral. Let \( u = \ln x \). Then, the derivative \( du = \frac{1}{x} dx \) can be rearranged to find \( dx \) in terms of \( du \): \( dx = x \, du \).

Substitute in the Variables

Replace \( \ln x \) with \( u \) and \( dx \) with \( x \, du \) in the integral. The original integral \( \int \frac{dx}{x(\ln x)^2} \) becomes \( \int \frac{x \, du}{x u^2} \). Simplifying this expression gives \( \int \frac{1}{u^2} \, du \).

Solve the Simplified Integral

The integral \( \int \frac{1}{u^2} \, du \) is straightforward. It can be rewritten as \( \int u^{-2} \, du \). The antiderivative of \( u^{-2} \) is \( -u^{-1} \), hence the result of the integral is \( -\frac{1}{u} + C \), where \( C \) is the constant of integration.

Substitute Back to Original Variable

Since we made a substitution in the beginning, we now need to substitute back using \( u = \ln x \). So, \( -\frac{1}{u} + C \) becomes \( -\frac{1}{\ln x} + C \). This is the indefinite integral of the original problem.

Key concepts

Substitution Method

When tackling indefinite integrals, the substitution method is a powerful tool. It simplifies complex expressions in integrals, making them easier to solve. In simple terms, it involves substituting part of the integral expression with a single variable.

This method is akin to solving a mystery: when you identify the key element and replace it with a simpler form, the solution becomes clearer. For example, in the given exercise, the expression \( \ln x \) within the integral is chosen as a substitution candidate. By letting \( u = \ln x \), we unravel the complexity and reveal the relationship \( du = \frac{1}{x} dx \). This new expression helps us transform and simplify the integral.

Here's a simple plan to use the substitution method effectively:

• Identify a part of the integrand to substitute, often something with a clear derivative.
• Express \( dx \) in terms of the new variable \( du \).
• Substitute both the expression and \( dx \) in the integral.
• The resulting integral should be simpler to solve.

Integration Techniques

Integration involves various techniques to find the antiderivative of complex expressions. Using the correct integration technique is essential.

Substitution is one of the fundamental techniques, especially useful when you notice that a part of the integrand can be rewritten as the derivative times some function. Here, recognizing derivatives hidden within the integrand is crucial for transforming the expression.

Other techniques include:

• Direct Integration: Suitable when the integrand is a basic function.
• Integration by Parts: Useful when handling products of functions.
• Partial Fraction Decomposition: Eases integrals of rational expressions.
• Trigonometric Substitution: Applied when integral involves trigonometric identities.

Matching the appropriate method with the integral type is key to simplifying and solving the problem effectively.

Antiderivative

An antiderivative is essentially the reverse process of differentiation. It refers to a function whose derivative gives back the original function being integrated.

In the context of this exercise, once we simplify the integral to \( \int \frac{1}{u^2} \, du \), we face the familiar task of finding its antiderivative. Rewriting this as \( \int u^{-2} \, du \), we look for a function whose derivative is \( u^{-2} \). Through our understanding of power rules, we know the antiderivative is \( -u^{-1} \). Always remember, the antiderivative also includes an arbitrary constant, up next!

The significance of antiderivatives lies in their wide application:

• They form the basis of solving differential equations.
• They aid in calculating net changes from rate functions.
• They are a key concept in the Fundamental Theorem of Calculus.

Constant of Integration

The constant of integration, often represented as \( C \), is an enigmatic yet crucial element in indefinite integrals.

When we integrate a function, we obtain an antiderivative that includes \( C \). This roadblock emerges because differentiation of constants results in zero, making them invisible in the derivative form. Therefore, to ensure we encapsulate all potential antiderivatives, we append \( C \) to our integral expression.

The constant of integration:

• Represents all possible vertical shifts of an antiderivative.
• Ensures inclusivity of all solutions, satisfying initial or boundary conditions.
• Highlights the indefinite nature of the integral, as an infinite set of functions can qualify as solutions.

In the exercise solution, \( -\frac{1}{\ln x} + C \) marks the general solution to the original indefinite integral, emphasizing the utmost importance of \( C \) in integrating functions.