Question

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

Short answer

The integral is \(-1 + \frac{1}{\ln(2)}\).

Step-by-step solution

Identify the type of integral

The integral given is a definite integral because it has limits from 2 to \( e \). It has the form \( \int_{2}^{e} \frac{dx}{(x \ln(x))^{2}} \).

Set up a substitution

For the integral \( \int \frac{dx}{(x \ln(x))^{2}} \), we choose \( u = \ln(x) \). Then, \( du = \frac{1}{x} dx \), or equivalently \( dx = x \, du \). Substitute \( x = e^u \) to convert \( dx = e^u \, du \).

Change the limits of integration

Substitute the limits of integration. When \( x = 2 \), \( u = \ln(2) \). When \( x = e \), \( u = 1 \). Thus, the new limits for \( u \) range from \( \ln(2) \) to \( 1 \).

Substitute and simplify the integral

Substitute the values into the integral:\[\int_{\ln(2)}^{1} \frac{e^u \, du}{((e^u) \cdot u)^2} = \int_{\ln(2)}^{1} \frac{du}{u^2}.\]This simplifies to \( \int_{\ln(2)}^{1} u^{-2} \, du \).

Integrate

Integrate \( u^{-2} \) with respect to \( u \):\[\int u^{-2} \, du = -u^{-1} = -\frac{1}{u}.\]

Evaluate the definite integral

Evaluate the integral from \( \ln(2) \) to \( 1 \):\[\left[ -\frac{1}{u} \right]_{\ln(2)}^{1} = -\frac{1}{1} - \left( -\frac{1}{\ln(2)} \right).\]This results in \( -1 + \frac{1}{\ln(2)} \).

Present the solution

The definite integral is:\[\int_{2}^{e} \frac{dx}{(x \ln(x))^{2}} = -1 + \frac{1}{\ln(2)}.\]

Key concepts

Definite Integral

A definite integral is a way to calculate the area under a curve between two specific points on the x-axis. In our problem, the definite integral is represented by \[ \int_{2}^{e} \frac{d x}{(x \ln (x))^{2}} \], where 2 and \( e \) are the limits of integration. By finding the definite integral, we quantify the cumulative value of the function from these two points.Some key features of definite integrals include:

• The limits of integration, denoted as the numbers at the bottom and top of the integral sign, determine where we start and end on the x-axis.
• The function inside the integral, often called the integrand, is what we're calculating the area under.
• The result of a definite integral is often a specific numerical value, representing the total accumulated change or area from start to end points.

This is different from an indefinite integral which has no specific bounds and includes a constant of integration.

Substitution Method

The substitution method is a powerful tool in integral calculus used to simplify complex integrals. In our example, we utilize substitution to transform the integrand into a more manageable form. We start by choosing a new variable, say \( u \), to replace a complicated part of the integrand.In our problem, we let \( u = \ln(x) \), which simplifies the integrand significantly. This choice leads to the expression \( du = \frac{1}{x} dx \), which helps in changing the differentials, resulting in \( dx = x \, du \). By performing these substitutions, we change the entire integral into a simpler format, as seen here:\[ \int \frac{dx}{(x \ln(x))^{2}} = \int \frac{du}{u^2} \].Why use substitution?

• It simplifies the integrand, making it easier to integrate.
• Provides a clearer path to solving otherwise complex integrals.
• Helps in recognizing standard forms of integration that we know how to solve.

Substitution effectively 'translates' the integral into more familiar territory.

Integration Limits

The concept of integration limits is crucial within definite integrals as they define the range over which we are calculating. When performing a substitution, as done here, it is essential to also change the limits of integration to reflect the new variable.In our exercise, the original limits were from 2 to \( e \). After substituting \( u = \ln(x) \), the limits of integration become:

• When \( x = 2 \), \( u = \ln(2) \).
• When \( x = e \), \( u = 1 \).

These new limits mean the integration now occurs from \( u = \ln(2) \) to \( u = 1 \).Why are integration limits important?

• The limits provide the boundaries for the area we are calculating.
• A correct transformation of limits ensures that we are calculating the appropriate section under the new curve represented by the substitution variable.
• It is crucial to change limits accordingly when the integrand is transformed, to maintain consistency in the calculation.

This enables us to accurately find the area even when the function's form changes due to substitution.