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Part 1
Express the integrand as a sum of partial fractions and evaluate the integral.
Integral from 7 to 14 StartFraction y Over y squared minus 3 y minus 4 EndFraction dy
Question content area bottom
Part 1
Express the integrand as a sum of partial fractions. 
StartFraction y Over y squared minus 3 y minus 4 EndFraction
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​(Simplify your answer. Use integers or fractions for any numbers in the​ expression.)

We are tasked with expressing the integrand as a sum of partial fractions and evaluating the integral:

\[
\int_7^{14} \frac{y}{y^2 - 3y - 4} \, dy
\]

### Step 1: Factor the denominator

First, we factor the quadratic in the denominator:

\[
y^2 - 3y - 4 = (y - 4)(y + 1)
\]

So the integral becomes:

\[
\int_7^{14} \frac{y}{(y - 4)(y + 1)} \, dy
\]

### Step 2: Express the fraction as a sum of partial fractions

We now express \(\frac{y}{(y - 4)(y + 1)}\) as a sum of partial fractions:

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

To solve for \(A\) and \(B\), multiply both sides of the equation by \((y - 4)(y + 1)\):

\[
y = A(y + 1) + B(y - 4)
\]

### Step 3: Solve for \(A\) and \(B\)

Expand both sides:

\[
y = A(y + 1) + B(y - 4) = A y + A + B y - 4B
\]

Combine like terms:

\[
y = (A + B)y + (A - 4B)
\]

Equating coefficients of \(y\) and the constant term:

- From the coefficient of \(y\): \(A + B = 1\)
- From the constant term: \(A - 4B = 0\)

### Step 4: Solve the system of equations

1. \(A + B = 1\)
2. \(A - 4B = 0\)

From equation (2), solve for \(A\):

\[
A = 4B
\]

Substitute this into equation (1):

\[
4B + B = 1
\]
\[
5B = 1 \quad \Rightarrow \quad B = \frac{1}{5}
\]

Now substitute \(B = \frac{1}{5}\) into \(A = 4B\):

\[
A = 4 \times \frac{1}{5} = \frac{4}{5}
\]

### Step 5: Write the partial fraction decomposition

Thus, we have:

\[
\frac{y}{(y - 4)(y + 1)} = \frac{4/5}{y - 4} + \frac{1/5}{y + 1}
\]

### Step 6: Evaluate the integral

Now we can write the integral as:

\[
\int_7^{14} \frac{y}{(y - 4)(y + 1)} \, dy = \int_7^{14} \left( \frac{4/5}{y - 4} + \frac{1/5}{y + 1} \right) \, dy
\]

This splits into two integrals:

\[
\frac{4}{5} \int_7^{14} \frac{1}{y - 4} \, dy + \frac{1}{5} \int_7^{14} \frac{1}{y + 1} \, dy
\]

Both of these integrals are straightforward natural logarithms:

\[
\frac{4}{5} \left[ \ln|y - 4| \right]_7^{14} + \frac{1}{5} \left[ \ln|y + 1| \right]_7^{14}
\]

### Step 7: Compute the values

First integral:

\[
\frac{4}{5} \left( \ln|14 - 4| - \ln|7 - 4| \right) = \frac{4}{5} (\ln 10 - \ln 3) = \frac{4}{5} \ln \left( \frac{10}{3} \right)
\]

Second integral:

\[
\frac{1}{5} \left( \ln|14 + 1| - \ln|7 + 1| \right) = \frac{1}{5} (\ln 15 - \ln 8) = \frac{1}{5} \ln \left( \frac{15}{8} \right)
\]

### Step 8: Combine the results

The final result is:

\[
\frac{4}{5} \ln \left( \frac{10}{3} \right) + \frac{1}{5} \ln \left( \frac{15}{8} \right)
\]

This is the value of the integral.

---

Would you like to explore more details on this? Here are some related questions:

1. How do we handle improper integrals with partial fractions?
2. What are other common methods to solve rational integrals?
3. Can partial fraction decomposition always be applied to integrals?
4. How can logarithmic rules simplify expressions involving integrals?
5. How do factored quadratics affect the form of partial fractions?

**Tip:** Partial fractions are particularly helpful in integrating rational functions where the degree of the numerator is less than that of the denominator.

Express the integrand as a sum of partial fractions and evaluate the integral: Integral from 7 to 14 of y / (y^2 - 3y - 4) dy

Learn how to express y / (y^2 - 3y - 4) as a sum of partial fractions and evaluate its integral from 7 to 14. This solution covers algebraic factoring, partial fraction decomposition, and logarithmic integration. Suitable for Grades 11-12 or early college math students.

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