Question

For the following exercises, find the definite or indefinite integral. $$ \int_{2}^{e} \frac{d x}{x \ln x} $$

Short answer

The integral is \(-\ln (\ln 2)\).

Step-by-step solution

Identify the Type of Integral

The given integral is a definite integral with limits from 2 to \( e \): \[ \int_{2}^{e} \frac{dx}{x \ln x} \]The function we are integrating is \( \frac{1}{x \ln x} \).

Use Substitution Method

To simplify the integral, let's use substitution. Set \( u = \ln x \), which means \( du = \frac{1}{x} dx \). This substitution transforms the integral bounds. - When \( x = 2 \), \( u = \ln 2 \)- When \( x = e \), \( u = \ln e = 1 \).The integral becomes:\[ \int_{\ln 2}^{1} \frac{1}{u} \, du \]

Integrate Using Basic Logarithmic Integral

The integral \( \int \frac{1}{u} \, du \) is a basic logarithmic integral whose solution is \( \ln |u| + C \). Therefore, the indefinite integral of our function is:\[ \ln |u| \] evaluated from \( u = \ln 2 \) to \( u = 1 \).

Evaluate at Boundaries

Plug the upper and lower bounds into the result from the previous step:\[ \ln |1| - \ln |\ln 2| \]Since \( \ln 1 = 0 \), the evaluation simplifies to:\[ 0 - \ln |\ln 2| = -\ln (\ln 2) \]

Final Answer

The definite integral evaluates to:\[ -\ln (\ln 2) \]

Key concepts

Definite Integral

A definite integral is a fundamental concept in calculus that calculates the net area under a curve over a certain interval. It is represented by the integral sign with upper and lower limits, which define the starting and ending points of the interval. In the problem we're discussing, the definite integral is \[ \int_{2}^{e} \frac{dx}{x \ln x} \]. These boundaries tell us that we are interested in the area under the curve of the function \( \frac{1}{x \ln x} \) from \( x = 2 \) to \( x = e \).Key points about definite integrals:

• The limits, or bounds, define the exact section of the graph we are examining.
• The result of a definite integral is a number that represents this area.
• In definite integrals, constants of integration are not needed, as the calculation provides a specific numerical result.

Substitution Method

The substitution method is a technique used in calculus to simplify the integration process. It often turns a difficult integral into a more manageable form. This involves substituting a part of the integral with a new variable, often denoted by \( u \), along with its differential, \( du \).Here's how it worked in our example:

• We set \( u = \ln x \), which simplifies the integrand, allowing easier integration.
• Then, we find \( du = \frac{1}{x} dx \), replacing \( dx \) in the integral.
• Substitution also requires changing the bounds from \( x \) values to \( u \) values. For \( x = 2 \), \( u = \ln 2 \), and for \( x = e \), \( u = 1 \).

This method reconfigures the original integral \( \int_{2}^{e} \frac{dx}{x \ln x} \) to the simpler form \( \int_{\ln 2}^{1} \frac{1}{u} \, du \), which is easier to evaluate.

Logarithmic Integral

The logarithmic integral is a special form of integration involving logarithmic functions. In our case, after substitution, the problem simplifies to the logarithmic integral \( \int \frac{1}{u} \, du \), which has a well-known antiderivative:\[ \int \frac{1}{u} \, du = \ln |u| + C \]A few points to remember:

• The natural logarithm \( \ln |u| \) comes into play when integrating \( 1/u \).
• This integral does not directly include the constant \( C \) when calculating definite integrals, as we are looking for a specific value between bounds.
• The computation focuses on evaluating \( \ln |u| \) at specific boundary points, then finding the difference between these evaluations.

Integral Bounds

Integral bounds, sometimes called limits of integration, are the starting and ending points in a definite integral. They meticulously define the portion of the curve and area under investigation. For our integral, these bounds were from 2 to \( e \).Steps in applying bounds:

• When using substitution, bounds must be adjusted to fit the new variable "\( u \)." In our case, the original bounds of 2 and \( e \) became \( \ln 2 \) and 1 respectively.
• The bounds help determine the specific values to evaluate the antiderivative at, leading to a numerical result.
• After evaluating the integral at these new bounds, subtract the value of the lower bound from the upper bound's value.
• As seen in the solution, you use these bounds to find \( \ln |1| - \ln |\ln 2| \), which ultimately simplifies to \( -\ln (\ln 2) \).

Integral bounds are critical as they ensure we are computing the exact area over our interval of interest.