If `int_(0)^(36) (1)/(2x+9)dx =log k`, is equal to

Options

A. 3
B. `9//2`
C. 9
D. 81

Answer

To solve the integral \(\int_{0}^{36} \frac{1}{2x + 9} \, dx\) and find \(k\) such that \(\int_{0}^{36} \frac{1}{2x + 9} \, dx = \log k\), we will follow these steps:

### Step 1: Set up the integral
Let
\[ I = \int_{0}^{36} \frac{1}{2x + 9} \, dx \]

### Step 2: Use the integration formula
We will use the formula for integrating \(\frac{1}{ax + b}\):
\[ \int \frac{1}{ax + b} \, dx = \frac{1}{a} \log |ax + b| + C \]
Here, \(a = 2\) and \(b = 9\).

### Step 3: Apply the formula to the integral
Applying the formula, we have:
\[ I = \frac{1}{2} \log |2x + 9| \bigg|_{0}^{36} \]

### Step 4: Evaluate the limits
Now we will evaluate the expression at the upper and lower limits:
\[ I = \frac{1}{2} \left( \log |2(36) + 9| - \log |2(0) + 9| \right) \]
Calculating the values:
\[ = \frac{1}{2} \left( \log |72 + 9| - \log |0 + 9| \right) \]
\[ = \frac{1}{2} \left( \log |81| - \log |9| \right) \]

### Step 5: Use the properties of logarithms
Using the property of logarithms \(\log a - \log b = \log \frac{a}{b}\):
\[ = \frac{1}{2} \log \left( \frac{81}{9} \right) \]
Calculating \(\frac{81}{9}\):
\[ = \frac{1}{2} \log(9) \]

### Step 6: Express \(9\) in terms of \(3\)
We can express \(9\) as \(3^2\):
\[ \log(9) = \log(3^2) = 2 \log(3) \]
Thus,
\[ I = \frac{1}{2} \cdot 2 \log(3) = \log(3) \]

### Step 7: Relate to \(k\)
From the problem, we have:
\[ I = \log k \]
Thus,
\[ \log k = \log(3) \]
This implies:
\[ k = 3 \]

### Final Answer
Therefore, the value of \(k\) is:
\[ \boxed{3} \]