Integral of \( \dfrac{\ln x}{x} \): How to Solve It

The integral \( \displaystyle \int \dfrac{\ln x}{x} \, dx \) is an example of integrating compund functions, a foundational skill in calculus. Further, $\ln x$ is a function that arises in various real-world contexts, as well as in classroom and test settings.

Solution

The antiderivative of \( \dfrac{\ln x}{x} \) with respect to \( x \) is:

\[\begin{align*} \displaystyle \int \dfrac{\ln x}{x} \, dx = \dfrac{(\ln x)^2}{2} + C, \end{align*}\]

where \( C \) is the constant of integration. This solution expresses the growth of \( \ln x \) squared, scaled by \( \dfrac{1}{2} \).

Explanation

To evaluate \( \displaystyle \int \dfrac{\ln x}{x} \, dx \), we use substitution. Let \( u = \ln x \), which gives \( du = \dfrac{1}{x} dx \). Substituting \( u \) and \( du \) transforms the integral into a simpler form:

\[\begin{align*} \displaystyle \int u \, du = \dfrac{u^2}{2} + C. \end{align*}\]

Returning to \( x \), substitute \( u = \ln x \) back in:

\[\begin{align*} \dfrac{(\ln x)^2}{2} + C. \end{align*}\]

This method shows how substitution can simplify integrals involving logarithmic functions, especially when the integrand includes both \( \ln x \) and a term related to \( x \).

Applications

This integral has applications in fields where the natural logarithm often shows up, such as:

• Information Theory: Calculating entropy, where expressions involving \( \ln x \) measure uncertainty.
• Economics: Modeling returns on investments, where logarithmic functions represent growth rates over time.
• Statistics: Used in distributions and data transformations.

Practice Problem with Solution

Problem: An economist models the income elasticity of demand as \( E(I) = \dfrac{\ln I}{I} \), where \( I \) is income in thousands of dollars. Calculate the total elasticity change from \( I = 1 \) to \( I = 10 \).

Solution: We need to find \( \displaystyle \int_1^{10} \dfrac{\ln I}{I} \, dI \). Using the formula \( \displaystyle \int \dfrac{\ln I}{I} \, dI = \dfrac{(\ln I)^2}{2} + C \), we evaluate the definite integral:

\[\begin{align*} \displaystyle \int_1^{10} \dfrac{\ln I}{I} \, dI = \left[\dfrac{(\ln I)^2}{2}\right]_1^{10}. \end{align*}\]

Substituting the bounds of integration gives

\[\begin{align*} \dfrac{(\ln 10)^2}{2} – \dfrac{(\ln 1)^2}{2} = \dfrac{(\ln 10)^2}{2}. \end{align*}\]

Since \( \ln 1 = 0 \), we simplify to \( \dfrac{(\ln 10)^2}{2} \), which represents the elasticity change over the income range.