https://file.aiquickdraw.com/mathbot/file/1728163852774znqiar73.jpg

The equation in the image is as follows:

\[
\int_0^1 \left( \frac{4}{5x+2} - \frac{K}{4x+9} \right) dx = \frac{4}{5} \ln \left( \frac{7}{2} \right) + 2 \ln \left( \frac{9}{13} \right)
\]

To find the value of \( K \), we need to solve the integral on the left-hand side and equate it to the right-hand side.

### Step 1: Evaluate the integral
The integral involves two rational functions:

\[
\int_0^1 \left( \frac{4}{5x+2} - \frac{K}{4x+9} \right) dx
\]

We can split this into two integrals:

\[
\int_0^1 \frac{4}{5x+2} dx - \int_0^1 \frac{K}{4x+9} dx
\]

#### First integral:
\[
\int_0^1 \frac{4}{5x+2} dx
\]

Let \( u = 5x + 2 \), so \( du = 5 dx \), which gives:

\[
\frac{4}{5} \int_2^7 \frac{1}{u} du = \frac{4}{5} \ln \left( \frac{7}{2} \right)
\]

#### Second integral:
\[
\int_0^1 \frac{K}{4x+9} dx
\]

Let \( v = 4x + 9 \), so \( dv = 4 dx \), which gives:

\[
\frac{K}{4} \int_9^{13} \frac{1}{v} dv = \frac{K}{4} \ln \left( \frac{13}{9} \right)
\]

### Step 2: Combine the integrals
We now combine both results:

\[
\frac{4}{5} \ln \left( \frac{7}{2} \right) - \frac{K}{4} \ln \left( \frac{13}{9} \right)
\]

### Step 3: Set the equation equal to the right-hand side
Equate this to the right-hand side of the equation:

\[
\frac{4}{5} \ln \left( \frac{7}{2} \right) + 2 \ln \left( \frac{9}{13} \right)
\]

### Step 4: Solve for \( K \)

Simplifying the right-hand side:

\[
2 \ln \left( \frac{9}{13} \right) = \ln \left( \frac{9^2}{13^2} \right) = \ln \left( \frac{81}{169} \right)
\]

Thus, the equation becomes:

\[
\frac{4}{5} \ln \left( \frac{7}{2} \right) - \frac{K}{4} \ln \left( \frac{13}{9} \right) = \frac{4}{5} \ln \left( \frac{7}{2} \right) + \ln \left( \frac{81}{169} \right)
\]

Since the terms involving \( \ln \left( \frac{7}{2} \right) \) cancel out, we are left with:

\[
-\frac{K}{4} \ln \left( \frac{13}{9} \right) = \ln \left( \frac{81}{169} \right)
\]

This simplifies to:

\[
-\frac{K}{4} \ln \left( \frac{13}{9} \right) = \ln \left( \frac{81}{169} \right)
\]

\[
-\frac{K}{4} = \frac{\ln \left( \frac{81}{169} \right)}{\ln \left( \frac{13}{9} \right)}
\]

Since \( \frac{\ln \left( \frac{81}{169} \right)}{\ln \left( \frac{13}{9} \right)} = 2 \), we find that:

\[
-\frac{K}{4} = 2
\]

Thus,

\[
K = -8
\]

### Conclusion:
The value of \( K \) is \( \boxed{-8} \).

---

Would you like more details or have any questions?

### Related Questions:
1. How can you apply substitution techniques in integrals more effectively?
2. What are some common logarithmic properties that are useful in solving integrals?
3. How would the answer change if the limits of integration were different?
4. How can you check if your integral evaluation is correct?
5. What is the significance of rational functions in calculus problems?

### Tip:
When dealing with logarithmic terms inside integrals, always use properties of logarithms to simplify the expressions before attempting to solve them.

Find the value of K by solving a definite integral involving rational functions and logarithms. This problem covers key concepts in calculus, including the fundamental theorem of calculus and substitution techniques. Suitable for undergraduate-level calculus students.

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