The antiderivative of \( \dfrac{1}{x} \) is a foundational result in calculus, and appears in a variety of contexts. Specifically, it reveals interesting properties of the natural logarithm in calculus.

Solution

The antiderivative of \( \dfrac{1}{x} \) with respect to \( x \) is:

\[\begin{align*} \displaystyle \int \dfrac{1}{x} \, dx = \ln |x| + C \end{align*}\]

where \( C \) is the constant of integration. The result follows from the fact that the derivative of \( \ln |x| \) is \( \dfrac{1}{x} \). This is a key relationship in calculus.

Why \( \ln |x| \)?

We use \( |x| \) to cover both positive and negative \( x \) values. This ensures the result is defined for all \( x \neq 0 \), as \( \dfrac{1}{x} \) is undefined at \( x = 0 \).

Applications

This integral shows up in calculations in fields such as:

• Finance:Calculating continuous compounding interest.
• Biology:Modeling population growth and decay.
• Statistics:Log-normal distributions.

Practice Problems and Solution

Problem 1: Compute \( \displaystyle \int \dfrac{1}{3x} \, dx \).

Solution: To integrate \( \dfrac{1}{3x} \), first factor out the constant \( \dfrac{1}{3} \):

\[\begin{align*} \displaystyle \int \dfrac{1}{3x} \, dx = \dfrac{1}{3} \displaystyle \int \dfrac{1}{x} \, dx. \end{align*}\]

Applying the antiderivative of \( \dfrac{1}{x} \), we get:

\[\begin{align*} \dfrac{1}{3} \ln |x| + C. \end{align*}\]

Thus, the solution is \( \dfrac{1}{3} \ln |x| + C \).

Problem 2: Find \( \displaystyle \int \dfrac{2}{x} \, dx \).

Solution: Again, factor out the constant \( 2 \):

\[\begin{align*} \displaystyle \int \dfrac{2}{x} \, dx = 2 \displaystyle \int \dfrac{1}{x} \, dx. \end{align*}\]

Using the antiderivative of \( \dfrac{1}{x} \):

\[\begin{align*} 2 \ln |x| + C. \end{align*}\]

So, the result is \( 2 \ln |x| + C \).

Problem 3: Evaluate \( \displaystyle \int_2^6 \dfrac{1}{x} \, dx \).

Solution: Here, we need the definite integral from \( x = 2 \) to \( x = 6 \):

\[\begin{align*} \displaystyle \int_2^6 \dfrac{1}{x} \, dx = \ln |x|\bigg|_2^6. \end{align*}\]

Substitute the limits:

\[\begin{align*} \ln |6| – \ln |2| = \ln 6 – \ln 2 = \ln \dfrac{6}{2} = \ln 3. \end{align*}\]

Thus, the answer is \( \ln 3 \), which represents the total change in logarithmic terms over this interval.