Problem 87 - University Calculus: Early Transcendentals

Question

Right, or wrong? Give a brief reason why. $$ \int \frac{-15(x+3)^{2}}{(x-2)^{4}} d x=\left(\frac{x+3}{x-2}\right)^{3}+C $$

Step-by-Step Solution

Answer

Right, as the derivative matches the integrand.

Step 1: Identify the Integral

The integral given is \( \int \frac{-15(x+3)^{2}}{(x-2)^{4}} \, dx \). We need to verify if integrating this expression results in \( \left(\frac{x+3}{x-2}\right)^{3} + C \).

Step 2: Rewrite the Integral in a Usable Form

Notice that the proposed solution hints at a substitution. Rewrite the integral \( \int \frac{-15(x+3)^{2}}{(x-2)^{4}} \, dx \) such that it resembles the derivative form of the proposed solution \( \left( \frac{x+3}{x-2} \right)^{3} \).

Step 3: Differentiate the Proposed Expression

Calculate the derivative of the function \( \left( \frac{x+3}{x-2} \right)^{3} \). Use the chain rule: if \( u = \frac{x+3}{x-2} \), then \( (u^3)' = 3u^2 \cdot u' \).

Step 4: Find the Derivative of the Inner Function

For \( u = \frac{x+3}{x-2} \), find \( \frac{d}{dx}\left( \frac{x+3}{x-2} \right) \). Use the quotient rule: \( \left( \frac{f}{g} \right)' = \frac{f'g - fg'}{g^2} \), where \( f = x+3 \) and \( g = x-2 \).

Step 5: Simplify the Derivatives

Compute: \( f' = 1 \), \( g' = 1 \). The derivative is \( \frac{(x-2) - (x+3)}{(x-2)^2} = \frac{-5}{(x-2)^2} \).

Step 6: Multiply and Simplify

Combine results from Steps 3 and 5: \( 3 \left( \frac{x+3}{x-2} \right)^{2} \cdot \frac{-5}{(x-2)^{2}} = \frac{-15(x+3)^{2}}{(x-2)^4} \).

Step 7: Conclusion

The derivative calculated in Step 6 matches the integrand. This means the integral \( \int \frac{-15(x+3)^{2}}{(x-2)^{4}} \, dx \) is correctly equal to \( \left( \frac{x+3}{x-2} \right)^{3} + C \).

Key Concepts

Integration Techniques

Integration is an essential part of calculus that involves finding a function that describes the accumulation of quantities. There are several techniques used to perform integration, each applicable in different scenarios.

• **Substitution:** Often used when the integrand involves a composite function. You can simplify the expression using a variable change, similar to reversing the chain rule used in differentiation.
• **Integration by Parts:** This technique comes from the product rule of differentiation. It's useful when dealing with products of functions that are not easily integrated separately.
• **Partial Fraction Decomposition:** Handy when the integrand is a rational function, as it allows breaking a complicated fraction into simpler fractions which are easier to integrate.

In this exercise, the method of substitution is indirectly used, as implied by the structure of the given expression. The integral \[\int \frac{-15(x+3)^{2}}{(x-2)^{4}} \, dx\]is matched to the derivative of a proposed function, showing how integration can reverse the process of differentiation. By recognizing the derivatives of components, we can assert that the integral is correct. This highlights the interconnected nature of integration and differentiation, often described as inverse operations.

Quotient Rule

The quotient rule is a technique used in calculus to find the derivative of a function given by one function divided by another. It's essential when dealing with rational functions, a common sight in calculus. The quotient rule states:
If you have a function in the form \( \frac{f(x)}{g(x)} \), its derivative \( \left( \frac{f}{g} \right)' \) is given by:\[\frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}\]In the exercise, the quotient rule is applied to differentiate the expression:\[\left( \frac{x+3}{x-2} \right)\]Here, \( f(x) = x+3 \) and \( g(x) = x-2 \). Differentiating these gives \( f'(x) = 1 \) and \( g'(x) = 1 \), so the derivative becomes:\[\frac{1(x-2) - (x+3)1}{(x-2)^2} = \frac{-5}{(x-2)^2}\]This derivative calculation is vital in verifying that our integral indeed yields the proposed solution. The quotient rule provides a systematic approach to handle complex rational derivatives, which is crucial for solving integration problems like the one in the exercise.

Chain Rule

The chain rule is a fundamental rule in differentiation, used to find the derivative of compositions of functions. It is particularly useful whenever a function is nested inside another. The chain rule states:
If a function \( y = f(g(x)) \) is composed in this way, the derivative \( y' \) is:\[f'(g(x)) \cdot g'(x)\]In this exercise, the derivative of the function\[\left( \frac{x+3}{x-2} \right)^{3}\]is a perfect example of applying the chain rule. The outer function is \( u^3 \), with \( u = \frac{x+3}{x-2} \). Differentiating using the chain rule, the derivative becomes:\[3u^2 \cdot u'\]We already computed \( u' = \frac{-5}{(x-2)^2} \), so substituting these gives us:\[3 \left( \frac{x+3}{x-2} \right)^2 \frac{-5}{(x-2)^2} = \frac{-15(x+3)^2}{(x-2)^4}\]This insight confirms that the integrand matches the derivative of our function. The chain rule is indispensable for dealing with integrals that involve complex nested functions, as it allows breaking down the problem into manageable pieces.