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The given integral is:

\[
\int_{2}^{\infty} \frac{dy}{y^2 + 2y - 3}
\]

To solve this, we will first simplify the denominator and then proceed with the integration.

### Step 1: Factor the quadratic expression

We can factor the denominator \( y^2 + 2y - 3 \) as:

\[
y^2 + 2y - 3 = (y + 3)(y - 1)
\]

So the integral becomes:

\[
\int_{2}^{\infty} \frac{dy}{(y + 3)(y - 1)}
\]

### Step 2: Partial fraction decomposition

We express the integrand as a sum of partial fractions:

\[
\frac{1}{(y + 3)(y - 1)} = \frac{A}{y + 3} + \frac{B}{y - 1}
\]

Multiplying both sides by \((y + 3)(y - 1)\) gives:

\[
1 = A(y - 1) + B(y + 3)
\]

Expanding the right-hand side:

\[
1 = A y - A + B y + 3B
\]

Group the terms:

\[
1 = (A + B)y + (-A + 3B)
\]

This gives us the system of equations:

\[
A + B = 0
\]
\[
-A + 3B = 1
\]

From the first equation, \( A = -B \). Substituting this into the second equation:

\[
-(-B) + 3B = 1
\]
\[
B + 3B = 1
\]
\[
4B = 1 \quad \Rightarrow \quad B = \frac{1}{4}
\]

Since \( A = -B \), we have \( A = -\frac{1}{4} \).

### Step 3: Rewrite the integrand

Now we can rewrite the integrand as:

\[
\frac{1}{(y + 3)(y - 1)} = \frac{-\frac{1}{4}}{y + 3} + \frac{\frac{1}{4}}{y - 1}
\]

So the integral becomes:

\[
\int_{2}^{\infty} \frac{-\frac{1}{4}}{y + 3} \, dy + \int_{2}^{\infty} \frac{\frac{1}{4}}{y - 1} \, dy
\]

This simplifies to:

\[
-\frac{1}{4} \int_{2}^{\infty} \frac{dy}{y + 3} + \frac{1}{4} \int_{2}^{\infty} \frac{dy}{y - 1}
\]

### Step 4: Integrate

Both integrals are of the form \(\int \frac{1}{x} dx = \ln |x|\). So we have:

\[
-\frac{1}{4} \ln |y + 3| \Bigg|_{2}^{\infty} + \frac{1}{4} \ln |y - 1| \Bigg|_{2}^{\infty}
\]

Now evaluate the limits:

For the first term:

\[
-\frac{1}{4} \left( \lim_{y \to \infty} \ln |y + 3| - \ln |2 + 3| \right) = -\frac{1}{4} \left( \lim_{y \to \infty} \ln y - \ln 5 \right)
\]

Since \(\lim_{y \to \infty} \ln y = \infty\), the first term tends to 0 as \( y \to \infty \), leaving:

\[
-\frac{1}{4} \ln 5
\]

For the second term:

\[
\frac{1}{4} \left( \lim_{y \to \infty} \ln |y - 1| - \ln |2 - 1| \right) = \frac{1}{4} \left( \lim_{y \to \infty} \ln y - \ln 1 \right)
\]

Since \(\ln 1 = 0\), the second term becomes:

\[
\frac{1}{4} \ln \infty = 0
\]

So the final result is:

\[
-\frac{1}{4} \ln 5
\]

Therefore, the value of the integral is:

\[
-\frac{1}{4} \ln 5
\]

Let me know if you want further details or have any questions!

---

Here are 5 related questions to explore further:
1. How does partial fraction decomposition work in general?
2. Can integrals of rational functions always be solved using partial fractions?
3. What is the behavior of logarithmic functions at infinity?
4. How does improper integration differ from definite integration?
5. How do you interpret the result of an integral in terms of area?

**Tip:** When factoring quadratic expressions in the denominator of integrals, always check for simple factorable forms first before considering other methods!

Evaluate the integral from 2 to infinity of 1/(y^2 + 2y - 3) dy.

Learn how to solve the improper integral from 2 to infinity of 1/(y^2 + 2y - 3) using partial fraction decomposition. This step-by-step solution covers calculus concepts such as factoring quadratic expressions and integrating using logarithms.

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