Problem 19 - Calculus: Early Transcendental Functions

Question

Evaluate the indicated integral. $$\int \frac{4}{x(\ln x+1)^{2}} d x$$

Step-by-Step Solution

Answer

The indefinite integral of \( \frac{4}{x(\ln x+1)^{2}} \) dx is \( -4(\ln{x} + 1)^{-1} + C \).

Step 1: Use Substitution

We use the substitution \( u = \ln{x} + 1 \). This leads to du = \( \frac{1}{x} \) dx. We can solve for dx, giving \( dx = x \) du. In our integral, \( x \) and dx will cancel, simplifying the integrand. The integral now becomes \( \int \frac{4}{u^{2}} \) du.

Step 2: Evaluate Integral

The given integral is a power rule problem. Using power rule for integrals, we integrate \( \int u^{-2} \) du. The integral of \( u^{-2} \) with respect to \( u \) is \( -u^{-1} + C \). When we multiply by 4, the integral becomes \( -4u^{-1} + C \).

Step 3: Back Substitute u

Now, we substitute \( u = \ln{x} + 1 \) back into the answer. This gives the final answer to be \( -4(\ln{x} + 1)^{-1} + C \).

Key Concepts

Integration Techniques

Integration is a fundamental concept in calculus that helps us find areas under curves and solve real-world problems involving accumulation. Several techniques can efficiently solve different types of integrals. Let's explore some commonly used ones:

• The Substitution Method simplifies integrals by changing variables.
• The Partial Fraction Decomposition breaks down complex rational expressions.
• The Integration by Parts is useful when the integrand is a product of functions.
• Trigonometric Integration applies when integrals involve trigonometric functions.

Each method can be the key to solving an integral. The choice of technique depends heavily on the form of the integrand. Understanding and practicing these techniques will enhance your problem-solving skills in calculus.

Substitution Method

The substitution method, also known as u-substitution, is a powerful technique that simplifies the process of integration by substituting a part of the integrand with a new variable.This method transforms a complicated integral into a more manageable one.
Here's how it generally works:

• First, select a part of the integrand to substitute with a new variable, say, \( u \).
• Then, compute the differential, \( du \), in terms of the original variable.
• Replace the parts of the integrand and the differential in terms of \( u \), simplifying the integral.
• Finally, integrate with respect to \( u \) and substitute back the original variable using the relationship defined by \( u \).

In our original exercise, we used \( u = \ln{x} + 1 \). By changing the variable, the integral simplifies significantly, making it easier to solve with the power rule for integrals.

Power Rule for Integrals

The power rule for integrals is a basic yet highly useful technique in calculus, applicable when integrating power functions. It's a direct extension of the power rule for derivatives.Here's the rule:

• For a function of the form \( x^n \), the integral is \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), where \( n eq -1 \).
• The constant \( C \) represents the constant of integration, accounting for indefinite integrals that could differ by a constant.

However, if \( n = -1 \), the integral becomes the natural logarithm, \( \int x^{-1} \, dx = \ln|x| + C \).

In our step-by-step solution, the function was \( u^{-2} \). Following the power rule, integrating \( u^{-2} \) gives \( -u^{-1} + C \), which becomes the term \( -4u^{-1} + C \) after multiplying by 4. Such applications emphasize the importance of this rule when dealing with polynomial-like functions in integrals.