Question

Find the integral by using the simplest method. Not all problems require integration by parts. $$ \left.\int \ln x d x \text { (Hint: } \int \ln x d x \text { is equivalent to } \int 1 \cdot \ln (x) d x .\right) $$

Short answer

The integral of \( \ln x \) is \( x \ln x - x + C \).

Step-by-step solution

Recognize the Structure of the Integral

The integral we are asked to solve is \( \int \ln x \, dx \). From the hint, we can rewrite this as \( \int 1 \cdot \ln(x) \, dx \), which gives us a clue that integration by parts may be required as it fits the form \( \int u \, dv \).

Choose Functions for Integration by Parts

For integration by parts, the formula is \( \int u \, dv = uv - \int v \, du \). We choose \( u = \ln x \) and \( dv = dx \). Our task is to compute \( du \) and \( v \).

Differentiate and Integrate Chosen Functions

Differentiate \( u = \ln x \) to obtain \( du = \frac{1}{x} \, dx \). Integrate \( dv = dx \) to obtain \( v = x \).

Apply Integration by Parts Formula

Substitute \( u \), \( du \), \( v \), and \( dv \) into the formula: \[ \int \ln x \, dx = x \ln x - \int x \frac{1}{x} \, dx \]. Simplify the integrand: \[ \int x \frac{1}{x} \, dx = \int 1 \, dx \].

Solve the Remaining Integral

Solve \( \int 1 \, dx = x \). Now substitute back into the equation from Step 4: \[ \int \ln x \, dx = x \ln x - x \].

Write the Final Answer with Constant of Integration

Include the constant of integration to express the most general form of the integral: \( \int \ln x \, dx = x \ln x - x + C \), where \( C \) is the constant of integration.

Key concepts

Integration by Parts

Integration by Parts is a versatile method used to solve integrals where products of functions are involved. This technique is an extension of the product rule for differentiation and is very helpful when we deal with integrals of the type \( \int u \, dv \). The core formula is:

• \[ \int u \, dv = uv - \int v \, du \]

To apply this method effectively, it is crucial to choose the correct \( u \) and \( dv \) because it can simplify our work significantly. Typically, we choose \( u \) as a function that becomes simpler when differentiated, and \( dv \) as the remaining part of the integral that is easy to integrate.

In the exercise \( \int \ln x \, dx \), choosing \( u = \ln x \) is strategic because the derivative \( \frac{1}{x} \) is simpler than the original function. By integrating \( dv = dx \), we get \( v = x \). This turns the integral into a manageable expression, ultimately leading us to \( x \ln x - \int 1 \, dx \), which simplifies further. This key step showcases the power of integration by parts in transforming complex integrals into simpler ones.

Logarithmic Functions

Logarithmic functions, such as \( \ln x \), are essential in calculus due to their unique properties and behavior. The function \( \ln x \) represents the natural logarithm, which is the logarithm to the base \( e\), where \( e \approx 2.71828 \). It is defined only for positive values of \( x \), making it a crucial function for numerous applications in growth, decay, and various other natural processes.

A significant aspect of working with logarithmic functions is their derivatives and integrals. Knowing that the derivative of \( \ln x \) is \( \frac{1}{x} \) helps us utilize integration by parts effectively in transformations. In integrals, \( \ln x \) often appears in contexts where its distinct logarithmic properties blend harmoniously with polynomial and other elementary functions.

• Logarithmic identities such as \( \ln(ab) = \ln a + \ln b \)
• The subtraction form \( \ln(a/b) = \ln a - \ln b \)

These identities can often simplify expressions involving logarithms, aiding in calculating integral solutions like in the current exercise.

Indefinite Integrals

Indefinite integrals are a fundamental concept in calculus, representing the inverse process of differentiation. It addresses the problem of finding a function whose derivative is given, and these integrals typically include a constant of integration, denoted as \( C \).

• The notation \( \int f(x) \, dx \) indicates an indefinite integral.
• The general solution resulting from integration helps describe a family of functions.

For example, integrating \( \int \ln x \, dx \) yields \( x \ln x - x + C \). Here, \( C \) is critical as it accounts for the constant difference between any two antiderivatives of a function.

The inclusion of \( C \) in the indefinite integral acknowledges that differentiation eliminates constants, thereby introducing ambiguity in the original function. By learning how to find indefinite integrals, you equip yourself with tools for solving a plethora of real-world problems, including physics, engineering, and economics, where initial conditions or particular values determine the unique solution from the general set.