Problem 35 - Calculus for Biology and Medicine

Question

(a) Use integration by parts to show that, for \(x>0\), $$ \int \frac{1}{x} \ln x d x=(\ln x)^{2}-\int \frac{1}{x} \ln x d x $$ (b) Use your result in (a) to evaluate $$ \int \frac{1}{x} \ln x d x $$

Step-by-Step Solution

Answer

\( \int \frac{1}{x} \ln x \, dx = \frac{1}{2}(\ln x)^2 + C \).

Step 1 - Identify the Functions for Integration by Parts

For integration by parts, choose \( u = \ln x \) and \( dv = \frac{1}{x} dx \). This makes \( du = \frac{1}{x} dx \) and \( v = \ln x \).

Step 2 - Apply Integration by Parts Formula

According to the integration by parts formula: \( \int u \, dv = uv - \int v \, du \). Substitute the chosen values: \( \int \frac{1}{x} \ln x \, dx = \ln x \cdot \ln x - \int \ln x \cdot \frac{1}{x} \, dx \).

Step 3 - Simplify the Equation

The equation from Step 2 simplifies to: \( \int \frac{1}{x} \ln x \, dx = (\ln x)^2 - \int \frac{1}{x} \ln x \, dx \). This verifies the equation given in part (a).

Step 4 - Solve the Equation for \( \int \frac{1}{x} \ln x dx \)

To find \( \int \frac{1}{x} \ln x \, dx \), add \( \int \frac{1}{x} \ln x \, dx \) to both sides to get: \( 2 \int \frac{1}{x} \ln x \, dx = (\ln x)^2 \).

Step 5 - Evaluate the Integral

Divide both sides of the equation by 2 to solve for the integral: \( \int \frac{1}{x} \ln x \, dx = \frac{1}{2}(\ln x)^2 + C \), where \( C \) is the integration constant.

Key Concepts

Integral Calculus

Integral calculus is a fundamental part of calculus, focusing on the concept of integration. Integration can be understood as the reverse process of differentiation and is used to compute areas under curves, among other applications. In the context of integration by parts, we employ a formula derived from the product rule for differentiation. This technique allows us to integrate products of functions by reducing them to simpler forms. Integration by parts is often expressed as:

• \( \int u \, dv = uv - \int v \, du \)

where \( u \) and \( v \) are functions of \( x \). This technique is particularly useful when the integral involves a product of two functions, as seen in the exercise where \( u = \ln x \) and \( dv = \frac{1}{x} dx \). By cleverly choosing these functions, we can simplify complex integrals into easier ones.

Logarithmic Functions

Logarithmic functions are an essential component of calculus, especially in integration. The logarithm \( \ln x \) is the inverse operation of the exponential function. When we talk about \( \ln x \), we're referring to the natural logarithm, which has the base \( e \) (approximately 2.718). This function grows very slowly compared to linear and polynomial functions, making it a unique feature in calculus.

Why is the natural logarithm so critical? Here are a few reasons:

• It transforms multiplication into addition: \( \ln(ab) = \ln a + \ln b \).
• It appears frequently in integration, as many functions can be expressed or approximated using logarithms.
• In the exercise at hand, the natural logarithm function \( \ln x \) is integrated using integration by parts to reveal interesting properties about how logarithms behave under integration.

Definite and Indefinite Integrals

Integrals in calculus come in two main types: definite and indefinite. Understanding the difference is crucial for solving problems effectively.

**Indefinite Integrals**: These are represented as \( \int f(x) \, dx \) and represent a family of functions rather than specific values. The result typically includes a constant \( C \), showing the infinite range of possible functions differing by a constant. In the given solution:

• We find the indefinite integral of \( \frac{1}{x} \ln x \).
• The result includes \( \frac{1}{2} (\ln x)^2 + C \), showing many functions can satisfy the integral.

**Definite Integrals**: These compute exact areas and have specific limits of integration, denoted by \( \int_{a}^{b} f(x) \, dx \). Unlike indefinite integrals, they do not include a constant and evaluate the function's net area between the specified bounds. While the exercise focuses on an indefinite integral, the concepts learned are foundational for evaluating definite integrals by applying limits to the antiderivative or integral result. This underpins real-world applications, such as calculating total accumulated quantities.