Problem 57 - AP Calculus with 8 Practice Tests 15th

Question

\(\int \frac{y-1}{y+1} d y=\) (A) \(y-2 \ln |y+1|+C\) (B) \(1-\frac{2}{y+1}+C\) (C) \(\ln \frac{|y|}{(y+1)^{2}}+C\) (D) \(1-2 \ln |y+1|+C\)

Step-by-Step Solution

Answer

The correct answer is (A) \( y-2 \ln |y+1|+C \).

Step 1: Simplify the Fraction

First, let's simplify the fraction \( \frac{y-1}{y+1} \). We notice that the numerator can be rewritten as \( (y+1) - 2 \), which simplifies the fraction: \[ \frac{(y+1) - 2}{y+1} = \frac{y+1}{y+1} - \frac{2}{y+1} = 1 - \frac{2}{y+1}. \]

Step 2: Set Up the Integral

Using the simplified expression from Step 1, our integral becomes: \[ \int \left(1 - \frac{2}{y+1}\right) dy. \] This can be separated into two separate integrals: \[ \int 1 \, dy - 2 \int \frac{1}{y+1} \, dy. \]

Step 3: Integrate the First Term

The first integral is straightforward: \[ \int 1 \, dy = y. \] This is a direct integration since the integral of a constant is the constant times the variable of integration.

Step 4: Integrate the Second Term

For the second integral, recall the integral of \( \frac{1}{u} \) is \( \ln |u| \). Therefore, \[ \int \frac{1}{y+1} \, dy = \ln |y+1|. \] Applying the constant factor of \(-2\), we get the result \(-2 \ln |y+1|.\)

Step 5: Combine the Results

Add the results from Steps 3 and 4: \[ y - 2 \ln |y+1| + C. \] This represents the integral in its final form, where \( C \) is the constant of integration.

Step 6: Match with Given Options

Compare our result with the provided choices. We find that option (A) matches our solution: \( y - 2 \ln |y+1| + C.\)

Key Concepts

Integration Techniques

Integration is a fundamental concept in calculus that allows us to find the antiderivative of a function. This process is crucial for determining the area under curves and solving many real-world problems. In this particular exercise, we utilize a simplification technique to make the integration process easier.

First, by simplifying the fraction \( \frac{y-1}{y+1} \), we rewrite it as \( 1 - \frac{2}{y+1} \). This step breaks down a potentially challenging fraction into simpler integrable parts.

The approach used here is known as decomposition, where we express a function as a sum of more basic functions, in this case, \(1\) and \(-\frac{2}{y+1}\). Once split, each portion can be integrated separately, a method particularly handy when dealing with rational functions like the one in this exercise.

Definite and Indefinite Integrals

In calculus, integrals can be classified as either definite or indefinite. An indefinite integral, like the one in our exercise, represents a family of functions whose derivative is the original function. It includes a constant of integration, denoted by \( C \), acknowledging that the antiderivative of a function is not unique.

Definite integrals, on the other hand, reflect the net area under a curve between specific limits. They result in a number, not a function. The limits dictate the interval on the x-axis over which the area is calculated.

For indefinite integrals, such as \( \int \left(1 - \frac{2}{y+1}\right) \, dy \), the main goal is to reverse the differentiation process. Each term within the integral is integrated separately, yielding \( y - 2 \ln |y+1| + C \). The constant \( C \) is crucial, representing any vertical shift of the function's graph.

Logarithmic Integration

Logarithmic integration is an important technique necessary when dealing with functions of the form \( \frac{1}{u} \). The integral of such a function is logarithmic, expressed as \( \ln |u| \).

In this exercise, when we integrate \( \frac{1}{y+1} \), this method is essential. The result is \( \ln |y+1| \), a common outcome when handling rational functions in integrals.

• Recognizing integrals that simplify to logarithms is crucial.
• The absolute value ensures the logarithm is defined for all values of \( y \).

Understanding the connection between differentiation and integration of logarithmic functions deepens comprehension on how these operations undo one another. Incorporating constant factors, such as \(-2\), require careful attention, as they directly influence the final result by multiplying the entire logarithmic term.