If `int_(0)^(40) (dx)/( 2x +1) = log k`, then the value of k is

Options

A. 3
B. `(9)/(2)`
C. 9
D. none of these

Answer

To solve the integral \( \int_{0}^{40} \frac{dx}{2x + 1} = \log k \), we will follow these steps:

### Step 1: Substitution
Let \( t = 2x + 1 \). Then, we differentiate to find \( dx \):
\[ dt = 2dx \quad \Rightarrow \quad dx = \frac{dt}{2} \]

### Step 2: Change the limits of integration
Now, we need to change the limits of integration according to our substitution:
- When \( x = 0 \):
\[ t = 2(0) + 1 = 1 \]
- When \( x = 40 \):
\[ t = 2(40) + 1 = 81 \]

### Step 3: Rewrite the integral
Now we can rewrite the integral in terms of \( t \):
\[ \int_{0}^{40} \frac{dx}{2x + 1} = \int_{1}^{81} \frac{1}{t} \cdot \frac{dt}{2} = \frac{1}{2} \int_{1}^{81} \frac{dt}{t} \]

### Step 4: Evaluate the integral
The integral \( \int \frac{dt}{t} \) is \( \log t \). Therefore, we have:
\[ \frac{1}{2} \int_{1}^{81} \frac{dt}{t} = \frac{1}{2} \left[ \log t \right]_{1}^{81} = \frac{1}{2} \left( \log 81 - \log 1 \right) \]
Since \( \log 1 = 0 \):
\[ = \frac{1}{2} \log 81 \]

### Step 5: Simplify using properties of logarithms
Using the property of logarithms \( \log a^b = b \log a \):
\[ \frac{1}{2} \log 81 = \log 81^{1/2} = \log 9 \]

### Step 6: Set the equation equal to \( \log k \)
From the problem statement, we have:
\[ \log k = \log 9 \]

### Step 7: Solve for \( k \)
Since the logarithms are equal, we can conclude:
\[ k = 9 \]

### Final Answer
Thus, the value of \( k \) is \( 9 \).

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