2

1
 
4y2 − 2y − 12
y(y + 2)(y − 3)
 dy

To solve the integral of the given expression:

\[
\int \frac{4y^2 - 2y - 12}{y(y + 2)(y - 3)} \, dy
\]

We will use **partial fraction decomposition** to break it into simpler fractions, which can then be integrated individually.

### Step 1: Partial Fraction Decomposition
We decompose the given rational function into the form:

\[
\frac{4y^2 - 2y - 12}{y(y + 2)(y - 3)} = \frac{A}{y} + \frac{B}{y + 2} + \frac{C}{y - 3}
\]

Now, multiply both sides by the denominator \( y(y + 2)(y - 3) \):

\[
4y^2 - 2y - 12 = A(y + 2)(y - 3) + B(y)(y - 3) + C(y)(y + 2)
\]

### Step 2: Expand and Solve for \( A \), \( B \), and \( C \)
Expand both sides:

\[
4y^2 - 2y - 12 = A(y^2 - y - 6) + B(y^2 - 3y) + C(y^2 + 2y)
\]

This simplifies to:

\[
4y^2 - 2y - 12 = A(y^2 - y - 6) + B(y^2 - 3y) + C(y^2 + 2y)
\]

Now collect like terms:

\[
4y^2 - 2y - 12 = (A + B + C)y^2 + (-A - 3B + 2C)y + (-6A)
\]

By comparing the coefficients of \( y^2 \), \( y \), and the constant term, we get the following system of equations:

1. \( A + B + C = 4 \)  (coefficient of \( y^2 \))
2. \( -A - 3B + 2C = -2 \)  (coefficient of \( y \))
3. \( -6A = -12 \)  (constant term)

### Step 3: Solve the System of Equations
From the third equation:

\[
-6A = -12 \quad \Rightarrow \quad A = 2
\]

Substitute \( A = 2 \) into the first and second equations:

1. \( 2 + B + C = 4 \quad \Rightarrow \quad B + C = 2 \)
2. \( -2 - 3B + 2C = -2 \quad \Rightarrow \quad -3B + 2C = 0 \)

Now, solve the system:

From the second equation, solve for \( C \):

\[
2C = 3B \quad \Rightarrow \quad C = \frac{3}{2}B
\]

Substitute this into \( B + C = 2 \):

\[
B + \frac{3}{2}B = 2 \quad \Rightarrow \quad \frac{5}{2}B = 2 \quad \Rightarrow \quad B = \frac{4}{5}
\]

Substitute \( B = \frac{4}{5} \) into \( C = \frac{3}{2}B \):

\[
C = \frac{3}{2} \times \frac{4}{5} = \frac{6}{5}
\]

### Step 4: Rewrite the Integral
Now that we have \( A = 2 \), \( B = \frac{4}{5} \), and \( C = \frac{6}{5} \), the integral becomes:

\[
\int \left( \frac{2}{y} + \frac{\frac{4}{5}}{y + 2} + \frac{\frac{6}{5}}{y - 3} \right) dy
\]

This can be split into separate integrals:

\[
2 \int \frac{1}{y} \, dy + \frac{4}{5} \int \frac{1}{y + 2} \, dy + \frac{6}{5} \int \frac{1}{y - 3} \, dy
\]

### Step 5: Integrate Each Term
Now, integrate each term:

\[
2 \ln |y| + \frac{4}{5} \ln |y + 2| + \frac{6}{5} \ln |y - 3| + C
\]

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

### Final Answer:
\[
\boxed{2 \ln |y| + \frac{4}{5} \ln |y + 2| + \frac{6}{5} \ln |y - 3| + C}
\]

Would you like any further details or have any questions?

Here are 5 related questions for further practice:
1. How do you perform partial fraction decomposition when the denominator has repeated factors?
2. What are the properties of logarithms that can simplify log expressions?
3. How do you solve integrals that involve irreducible quadratic factors?
4. Can you explain how to integrate rational functions with a higher degree in the numerator than the denominator?
5. How can we apply integration techniques to solve real-world problems in physics?

**Tip**: Always simplify the algebra before proceeding with partial fractions—it makes solving systems of equations much easier!

Solve the integral: \( \int \frac{4y^2 - 2y - 12}{y(y + 2)(y - 3)} \, dy \)

Learn how to solve the integral of \( \frac{4y^2 - 2y - 12}{y(y + 2)(y - 3)} \) using partial fraction decomposition. This problem involves breaking down a rational function into simpler fractions and integrating with logarithms. Suitable for undergraduate calculus students.

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