To find the integral of \(\frac{y - 1}{y + 1} \, \mathrm{d}y\), we can use algebraic manipulation to simplify the integrand.

First, let's rewrite the integrand:

\[ \frac{y - 1}{y + 1} \]

We can split this fraction into two simpler fractions:

\[ \frac{y - 1}{y + 1} = \frac{y + 1 - 2}{y + 1} = \frac{y + 1}{y + 1} - \frac{2}{y + 1} \]

Simplifying the terms, we get:

\[ \frac{y - 1}{y + 1} = 1 - \frac{2}{y + 1} \]

Now, we can integrate each term separately:

\[ \int \left(1 - \frac{2}{y + 1}\right) \, \mathrm{d}y = \int 1 \, \mathrm{d}y - \int \frac{2}{y + 1} \, \mathrm{d}y \]

The integral of 1 with respect to \(y\) is:

\[ \int 1 \, \mathrm{d}y = y \]

The integral of \(\frac{2}{y + 1}\) can be found using the substitution method. Let \( u = y + 1 \), then \( \mathrm{d}u = \mathrm{d}y \):

\[ \int \frac{2}{y + 1} \, \mathrm{d}y = 2 \int \frac{1}{u} \, \mathrm{d}u = 2 \ln|u| = 2 \ln|y + 1| \]

Putting it all together:

\[ \int \frac{y - 1}{y + 1} \, \mathrm{d}y = y - 2 \ln|y + 1| + C \]

where \(C\) is the constant of integration.

Would you like more details or have any questions about this?

Here are five related questions you might consider:

1. How do you integrate rational functions in general?
2. What is the method of partial fractions, and how is it used in integration?
3. Can you provide examples of integrals requiring substitution?
4. How does the logarithmic function arise in integration?
5. What are some techniques for integrating complex rational expressions?

**Tip:** When integrating rational functions, consider simplifying the integrand through algebraic manipulation or substitution to make the integration process easier.

Learn how to find the integral of (y - 1) / (y + 1) with a detailed step-by-step solution. This guide covers algebraic manipulation, substitution method, and integration techniques. Suitable for college-level calculus students.

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