Problem 315 - Calculus Volume 1

Question

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

Step-by-Step Solution

Answer

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

Step 1: Identify the Integral

We are given the definite integral \( \int_{2}^{e} \frac{d x}{(x \ln (x))^{2}} \). This is an improper integral because it involves a division by \((x \ln(x))^{2}\). Our goal is to find its exact value between the limits 2 and \(e\).

Step 2: Substitute to Simplify the Integrand

Let \( u = \ln(x) \). Then, \( du = \frac{1}{x} dx \) or \( dx = x du \). Our expression \( x = e^u \) and the limits change from \( x=2 \) to \( u = \ln(2) \) and \( x=e \) to \( u=1 \). This gives us \( x \ln(x) = ue^u \), so the integrand becomes \( \int \frac{x}{(ue^u)^2} dx = \int \frac{1}{u^2} du \).

Step 3: Calculate the Integral with Respect to \( u \)

The integral \( \int \frac{1}{u^2} du \) is equivalent to \( \int u^{-2} du \). Using the power rule for integration, \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \), we find that \( \int u^{-2} du = \frac{-1}{u} + C \).

Step 4: Apply the Definite Limits

Now substitute back the limits for \( u \). Our integral becomes \( \left[\frac{-1}{u}\right]_{u=\ln(2)}^{u=1} = \left(-1\right)\left(\frac{1}{1} - \frac{1}{\ln(2)}\right) \). Simplify this to get \( 1 - \frac{1}{\ln(2)} \).

Step 5: Evaluate the Expression

Simplify \( 1 - \frac{1}{\ln(2)} \) to find the definite integral value. Numerically, this results in \( 1 - \frac{1}{\ln(2)} \), which cannot be simplified exactly to a rational number.

Key Concepts

Improper Integral

An improper integral is a type of integral where either the interval of integration or the function itself approaches infinity or has a point of discontinuity. In our exercise, the given integral \( \int_{2}^{e} \frac{d x}{(x \ln(x))^{2}} \) is considered improper. This is because the denominator \((x \ln(x))^{2}\) becomes undefined at specific points due to possible division by zero or infinite values. Analyzing such integrals involves checking their convergence, ensuring that the value of the integral does not diverge or become infinite. By correctly identifying this as an improper integral, we can apply special techniques to evaluate it, knowing that the boundaries or the function might exhibit behavior leading towards infinity or discontinuity.

Substitution Method

The substitution method is a crucial technique used to simplify integrals. It replaces a complicated part of the integrand with a new variable, making the integration process more manageable. In our example, we perform a substitution by letting \( u = \ln(x) \). This results in \( du = \frac{1}{x} dx \) or \( dx = x \, du \). By expressing \( x \) in terms of \( u \) as \( x = e^u \), we transform the integral's limits from \( x = 2 \) to \( u = \ln(2) \), and from \( x = e \) to \( u = 1 \). Consequently, the integrand \( \frac{x}{(ue^u)^2} \) simplifies to \( \frac{1}{u^2} \) in terms of \( u \). This substitution drastically simplifies the integral, making the solution attainable using standard integration rules.

Power Rule for Integration

The power rule for integration states that for any real number \( n eq -1 \), the integral \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \). In our problem, we use this rule to solve \( \int u^{-2} \, du \).Calculating this integral, we apply the power rule by setting \( n = -2 \). Here, the result is \( \int u^{-2} du = \frac{-1}{u} + C \). The constant of integration \( C \) is omitted when we evaluate definite integrals, which have specified upper and lower bounds. Applying this rule makes finding solutions to specific types of integrals straightforward and helps in algebraic manipulation where base functions are powers of \( u \).

Natural Logarithm

The natural logarithm, denoted \( \ln(x) \), is the logarithmic function with base \( e \), where \( e \) is approximately 2.71828. It's fundamental in calculus due to its simple derivative: \( \frac{d}{dx} \ln(x) = \frac{1}{x} \). The inverse function of the exponential function \( e^x \), natural logarithms simplify complex multiplicative relationships into additive ones.In solving integrals, the natural logarithm functions as part of substitution techniques. As illustrated in our exercise, choosing \( u = \ln(x) \) converted our problem into a simpler form. The natural logarithm is also key in dealing with exponential growth or decay processes, common in mathematical modeling of natural phenomena and financial computations.