Question

In the following exercises, integrate using the indicated substitution. $$ \int \frac{y-1}{y+1} d y ; u=y+1 $$

Short answer

Integral solution: \( y + 1 - 2 \ln |y+1| + C \).

Step-by-step solution

Identify the Substitution

The exercise provides us with the substitution \( u = y + 1 \). This will help us transform the integral into a simpler form by rewriting it in terms of \( u \) instead of \( y \).

Differentiate the Substitution

Differentiate \( u = y + 1 \) with respect to \( y \) to find the differential \( d y \). This results in \( d u = d y \). Hence, \( d y = d u \).

Express the Integral in Terms of u

Substitute \( y = u - 1 \) and \( d y = d u \) into the integral. This changes the integral from \( \int \frac{y-1}{y+1} d y \) to \( \int \frac{(u-1)-1}{u} d u = \int \frac{u-2}{u} d u \).

Split the Integral

Break down the integral \( \int \frac{u-2}{u} d u \) into two separate integrals: \( \int \frac{u}{u} d u - \int \frac{2}{u} d u \). This simplifies to \( \int 1 \, d u - 2 \int \frac{1}{u} d u \).

Integrate Each Part

Integrate both parts separately: \( \int 1 \, d u = u \) and \( -2 \int \frac{1}{u} d u = -2 \ln |u| \).

Write the General Solution

Combine the results from Step 5 to write the solution: \( u - 2 \ln |u| + C \), where \( C \) is the constant of integration.

Substitute Back in Terms of y

Finally, substitute back \( u = y + 1 \) to express the solution in terms of \( y \). The final result is \( y + 1 - 2 \ln |y+1| + C \).

Key concepts

Substitution Method

The substitution method is a technique used to simplify complex integrals by introducing a new variable. This makes the integration process more straightforward.
This method is particularly handy when dealing with integrals that contain nested or composite functions. By substituting one part of the function with a variable, we can rewrite the integral in a form that is easier to work with.
To effectively apply the substitution method:

• Identify the part of the integral to substitute (often a difficult or repetitive component).
• Introduce a substitution variable (\( u \)) and set it equal to the chosen component.
• Differentiate to find the differential relationship (\( du \)).
• Rewrite the entire integral in terms of the new variable.

This transformation often results in a simpler integral, which can then be integrated using basic rules.

Definite and Indefinite Integrals

Integrals come in two main types: definite and indefinite. Indefinite integrals represent families of antiderivatives without specified limits, characterized by the plus constant (\( + C \)).
Definite integrals, however, have specific upper and lower limits, providing a numeric result which represents the area under the curve.
Key points to understand:

• Indefinite integrals include a constant because they represent a family of functions. This constant matters when verifying initial conditions in applications.
• For definite integrals, you evaluate the antiderivative at both limits and take the difference, yielding a specific, finite value.
• In substitution for definite integrals, you must adjust limits based on the change of variables, if needed.

Both forms are essential for various applications, from computing areas to solving differential equations.

Integration

Integration is the reverse process of differentiation. While differentiation computes the rate of change, integration accumulates values over an interval.
It's an indispensable tool for calculating areas, volumes, and other quantities represented by continuous functions.
When you integrate a function, you're essentially finding its antiderivative. This involves determining which function, when differentiated, results in the integrand.
Techniques like substitution, integration by parts, and partial fractions make integration manageable even for complex functions.
Keep in mind:

• Always include the constant of integration (\( + C \)) for indefinite integrals.
• Verify results by differentiating back to check if the original function can be recovered.
• Know common integration rules and formulas for efficiency.

Through integration, you gain insights into the cumulative aspects of mathematical processes.

Antiderivatives

An antiderivative of a function is another function whose derivative is the original function.
Finding antiderivatives is crucial because it forms the basis of integration.
Consider it the backwards step of differentiation. For instance, if \( F(x) \) is an antiderivative of a function \( f(x) \), then \( F'(x) = f(x) \).Determining antiderivatives allows us to integrate functions and solve problems related to areas under curves or solving differential equations.
Important aspects include:

• Antiderivatives are not unique. They form a family of functions differing by a constant (\( + C \)).
• Verification is straightforward—differentiate your result to confirm it matches the original function.
• The process often involves reversing known differentiation rules.

Mastering antiderivatives simplifies the integration process and aids in various mathematical analyses.