Problem 61 - Calculus for Biology and Medicine

Question

The integral $$ \int \ln x d x $$ can be evaluated in two ways. (a) Write \(\ln x=1 \cdot \ln x\) and use integration by parts to evaluate the integral. (b) Use the substitution \(u=\ln x\) and integration by parts to evaluate the integral.

Step-by-Step Solution

Answer

The integral \( \int \ln x \, dx \) evaluates to \( x \ln x - x + C \) using both integration by parts and substitution methods.

Step 1: Identify the Integration Method for Part (a)

For part (a), we see that \( \ln x \) can be rewritten as \( 1 \cdot \ln x \). We will use integration by parts, which states that \( \int u \, dv = uv - \int v \, du \). Here, we choose \( u = \ln x \) and \( dv = dx \).

Step 2: Compute Derivatives and Integrals for Part (a)

Differentiating \( u = \ln x \) gives \( du = \frac{1}{x} \, dx \). Integrating \( dv = dx \) gives \( v = x \).

Step 3: Apply Integration by Parts for Part (a)

Using the integration by parts formula, we have: \[ \int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} \, dx = x \ln x - \int dx. \] This simplifies to \( x \ln x - x + C \), where \( C \) is the integration constant.

Step 4: Identify Substitution and Integration Method for Part (b)

For part (b), we use the substitution \( u = \ln x \) which implies \( du = \frac{1}{x} \, dx \) and \( x = e^u \). The integral becomes \( \int u \, e^u \, du \).

Step 5: Apply Integration by Parts for Part (b)

Let \( u = u \) and \( dv = e^u \, du \). Then, \( du = du \) and \( v = e^u \). Applying integration by parts yields: \[ \int u \, e^u \, du = u \, e^u - \int e^u \, du = u \, e^u - e^u + C. \] Substituting back \( u = \ln x \) and \( e^u = x \) gives \( x \ln x - x + C \).

Step 6: Compare Results

Both methods provided the same result: \( x \ln x - x + C \). This confirms the correctness of the solution via different approaches.

Key Concepts

Integration by Parts

Integration by parts is a powerful technique used in calculus to integrate products of functions. This method is based on the product rule for differentiation and is particularly useful in cases where direct integration is difficult. The formula for integration by parts is:\[\int u \, dv = uv - \int v \, du\]Here's how it works:

• First, choose which part of the integrand to assign to \( u \) and \( dv \). A common mnemonic is "LIATE" which stands for Logarithm, Inverse, Algebraic, Trigonometric, and Exponential. This helps in deciding which part of the integrand to differentiate and which to integrate.
• Differentiate \( u \) to find \( du \), and integrate \( dv \) to find \( v \).
• Substitute these into the integration by parts formula.
• Simplify and solve the resulting integral.

In our example, we use \( u = \ln x \), making \( dv = dx \). Differentiating \( u \) gives \( du = \frac{1}{x} \, dx \), and integrating \( dv \) gives \( v = x \). Following these steps leads us to the result \( x \ln x - x + C \). This systematic method helps break down complex integrals into simpler parts.

Integration Techniques

Mastering integration techniques is crucial for tackling a wide range of calculus problems. Various techniques can handle different forms of integral challenges. Some key integration techniques include:

• Integration by Parts: Useful for products of functions, as demonstrated by \( \int \ln x \, dx \).
• Substitution Method: Involves changing the variable to simplify the integrand, often used with trigonometric and exponential functions.
• Trigonometric Substitution: Ideal for integrands involving \( \sqrt{a^2 - x^2} \), converting them into trigonometric functions.
• Partial Fraction Decomposition: Applies to rational functions, breaking them into simpler fractions for easier integration.

Choosing the right technique depends on the form of the integrand and often requires a mix of algebraic manipulation and strategic thinking. Each method has its own set of rules and is well-suited for specific scenarios, making practice and experience essential for identifying the best approach.

Substitution Method

The substitution method is a fundamental technique in calculus for simplifying integrals. It involves changing variables to transform a complex integral into a more manageable form. Here's how you can apply substitution effectively:

• Choose a substitution that simplifies the integrand. Typically, this involves substituting a variable that appears repeatedly or affects multiple parts of the equation.
• Express the differential \( dx \) in terms of the new variable, \( du \). This will usually involve finding \( du \) as a derivative of your chosen substitution.
• Replace the original variable and differential in the integral with your substitutions. This transforms the integral into the new variable, often making it easier to solve.
• Integrate with respect to the new variable \( u \), and then substitute back the original variable to finish the problem.

In our exercise, setting \( u = \ln x \) and \( du = \frac{1}{x} \, dx \), with \( x = e^u \), converts the integral into \( \int u \, e^u \, du \). This simplifies the integration process and, when solved, results in the same integral solution \( x \ln x - x + C \). Substitution is a versatile and essential skill, particularly useful for complex expressions that resist more straightforward techniques.