Problem 49 - University Calculus: Early Transcendentals

Question

Evaluate the integrals. $$\int \frac{x}{\left(x^{2}-4\right)^{3}} d x$$

Step-by-Step Solution

Answer

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

Step 1: Identify the Method of Integration

To solve this integral, notice that the function inside the integrand is a composition involving a power of a binomial, \( (x^2 - 4)^3 \). This is a suitable scenario for substitution. We'll use the substitution method.

Step 2: Perform the Substitution

Let \( u = x^2 - 4 \). Then, the derivative \( du = 2x \, dx \) can be rearranged to \( x \, dx = \frac{1}{2} du \). Substitute these into the integral.

Step 3: Substitute and Simplify

Replace \( x^2 - 4 \) with \( u \) and \( x \, dx \) with \( \frac{1}{2} du \). The integral becomes: \[ \int \frac{x}{(x^2 - 4)^3} \ dx = \int \frac{1}{2u^3} \, du = \frac{1}{2} \int u^{-3} \, du \]

Step 4: Integrate with Respect to \( u \)

Integrate \( \frac{1}{2} \int u^{-3} \, du \). The antiderivative of \( u^{-3} \) is \( \frac{u^{-2}}{-2} \) or \( -\frac{1}{2}u^{-2} \). Thus, the integral becomes: \[ \frac{1}{2} \left( -\frac{1}{2u^2} \right) = -\frac{1}{4u^2} + C \] , where \( C \) is the constant of integration.

Step 5: Back-Substitute \( u \)

Replace \( u \) back with \( x^2 - 4 \). The expression \( -\frac{1}{4u^2} \) becomes \( -\frac{1}{4(x^2 - 4)^2} \). Consequently, the integral is: \[ -\frac{1}{4(x^2 - 4)^2} + C \]

Key Concepts

Substitution Method

The substitution method is a powerful tool in calculus, designed to simplify complex integrals, making them more manageable to evaluate. It involves replacing the integrand with a new variable, usually denoted as \( u \), to transform the original integral into an easier form. The key steps for employing substitution are as follows:

• Identify a Substitution: Look for a part of the integrand that, if replaced, will simplify the integral. In this case, we noticed the binomial \((x^2 - 4)^3\) and chose \( u = x^2 - 4 \).


• Calculate the Derivative: Compute the derivative of the chosen substitution, \( du = 2x \, dx \), and express \( x \, dx \) as a function of \( du \). Here, \( x \, dx = \frac{1}{2} du \).


• Replace and Simplify: Substitute \( u \) and \( du \) into the integral to transform it. Our initial integral \( \int \frac{x}{(x^2 - 4)^3} \, dx \) simplifies to \( \int \frac{1}{2u^3} \, du \).

This method is particularly helpful because it reduces the complexity of integration by transforming a more complicated expression into something more standard and recognizable.

Definite and Indefinite Integrals

When dealing with integrals, it is essential to distinguish between definite and indefinite integrals. Each type serves a unique purpose and provides different information.

• Indefinite integrals: These represent a family of functions and include the constant of integration \( C \). They express the general form of antiderivatives. For example, integrating \( \int u^{-3} \, du \) results in \( -\frac{1}{2}u^{-2} + C \).


• Definite integrals: These provide the actual value of an area under a curve between two bounds, \([a, b]\). It results in a real number and does not include \( C \) because limits are involved in the calculation. Although our example is an indefinite integral, the substitution method can also be used for definite integrals with respective limits exchanged for \( u \).

Understanding the difference between these types of integrals is fundamental in calculus as they allow us to compute areas and evaluate antiderivatives across a wide range of functions.

Calculus Problems

Calculus problems vary widely, but typically involve differentiation or integration to solve. Mastering these concepts is crucial for understanding change and area in mathematical contexts.

• Problem Identification: Begin by identifying the part of the calculus problem you are dealing with: differentiation (solving derivatives) or integration (solving integrals). Our focus is on integration here.


• Choosing Techniques: Depending on the structure of the function, choose the suitable method to integrate. This could be substitution, integration by parts, partial fractions, etc. For our function, substitution was most appropriate.


• Apply and Simplify: Apply the chosen technique, simplify, and solve. Integration often requires iterative refinement and step-by-step simplification as shown in our worked solution.

Each calculus problem you encounter builds your understanding, equipping you with tools to tackle complex mathematical challenges using integration or differentiation techniques. As demonstrated in our example, persistence not only aids in solving book problems but prepares you for practical applications in fields like physics, engineering, and beyond.