Problem 17 - AP Calculus with 8 Practice Tests 15th

Question

\(\int \frac{(x-2)^{3}}{x^{2}} d x=\) (A) \(\frac{(x-2)^{4}}{4 x^{2}}+C\) (B) \(\frac{x^{2}}{2}-6 x+6 \ln |x|-\frac{8}{x}+C\) (C) \(\frac{x^{2}}{2}-3 x+6 \ln |x|+\frac{4}{x}+C\) (D) \(\frac{x^{2}}{2}-6 x+12 \ln |x|+\frac{8}{x}+C\)

Step-by-Step Solution

Answer

(D) \(\frac{x^2}{2} - 6x + 12 \ln |x| + \frac{8}{x} + C\)

Step 1: Rewrite the integrand

Rewrite \( \frac{(x-2)^3}{x^2} \) as \( \frac{x^3 - 6x^2 + 12x - 8}{x^2} \). Expand \((x-2)^3\) to get \(x^3 - 6x^2 + 12x - 8\) and then divide each term by \(x^2\).

Step 2: Simplify the integrand

Divide each term in the polynomial by \(x^2\) to get: \(x - 6 + \frac{12}{x} - \frac{8}{x^2}\).

Step 3: Integrate term-by-term

Integrate each term separately:1. \(\int x \, dx = \frac{x^2}{2}\)2. \(\int -6 \, dx = -6x\)3. \(\int \frac{12}{x} \, dx = 12 \ln |x|\)4. \(\int -\frac{8}{x^2} \, dx = 8 \cdot \left(-\frac{1}{x}\right) = \frac{8}{x}\).

Step 4: Combine the results

Combine all the integrals to form the full antiderivative: \(\frac{x^2}{2} - 6x + 12 \ln |x| + \frac{8}{x} + C\).

Step 5: Select the correct answer

Compare this result with the given options. It matches option (D): \(\frac{x^2}{2} - 6x + 12 \ln |x| + \frac{8}{x} + C\).

Key Concepts

Polynomial Division

Polynomial Division is a crucial technique when working with integrals involving rational expressions. In this exercise, we have a fraction where the numerator has a polynomial expression and the denominator is a simple polynomial, specifically a monomial.
To simplify the integral of such a fraction, we begin by expanding the polynomial in the numerator. Here, \( (x-2)^3 \) is expanded to \( x^3 - 6x^2 + 12x - 8 \) to enable division by \( x^2 \). During polynomial division, each term of the expanded numerator is divided by the denominator individually, which results in the simplified expression: \( x - 6 + \frac{12}{x} - \frac{8}{x^2} \).
This simplification step is essential as it makes the integration process more straightforward by breaking a complex fraction into simpler terms that are easier to integrate individually.

Antiderivatives

An antiderivative, often called the indefinite integral, is the reverse process of differentiation. Finding antiderivatives is a key component of solving integrals, as it involves determining a function whose derivative yields the given function. In this context, we seek to find the antiderivative of each term obtained from the polynomial division.
For the simplified integrand \( x - 6 + \frac{12}{x} - \frac{8}{x^2} \):

• The antiderivative of \( x \) is \( \frac{x^2}{2} \).
• The antiderivative of a constant, like \( -6 \), is \( -6x \).
• For \( \frac{12}{x} \), the antiderivative is \( 12 \ln |x| \).
• The term \( -\frac{8}{x^2} \) integrates to \( \frac{8}{x} \).

Combining these results, we obtain the full antiderivative, or the indefinite integral, which is: \( \frac{x^2}{2} - 6x + 12 \ln |x| + \frac{8}{x} + C \), where \( C \) is the constant of integration representing any constant added to the solution.

Definite and Indefinite Integrals

Understanding the difference between definite and indefinite integrals is fundamental in integration techniques. An indefinite integral, like the one we found in this exercise, represents a family of functions and is expressed with a constant of integration \( C \). It does not compute specific values, but rather, it finds a general expression for the antiderivative of a function.
In contrast, a definite integral has limits of integration and calculates the exact area under a curve between two points, resulting in a number rather than a function. For instance, if we were to find the area under the curve \( y = x - 6 + \frac{12}{x} - \frac{8}{x^2} \) from \( a \) to \( b \), we would evaluate the definite integral with those limits:
\[\int_{a}^{b} \left( x - 6 + \frac{12}{x} - \frac{8}{x^2} \right) \, dx \] would produce a specific numerical result representing that area. Understanding these concepts helps in applying the right technique based on the problem requirements.