Choose the correct Answer of the Following Questions : `int_0^1 1/(2x-3) dx` is equal to

Options

A. 1/2 log3
B. log3
C. -1/2 log 3
D. 2log3

Answer

To solve the integral \( \int_0^1 \frac{1}{2x-3} \, dx \), we will use the method of substitution. Here are the detailed steps:

### Step 1: Substitution
Let \( z = 2x - 3 \). Then, we differentiate both sides to find \( dx \):
\[ dz = 2 \, dx \implies dx = \frac{dz}{2} \]

### Step 2: Change the limits of integration
Now, we need to change the limits of integration from \( x \) to \( z \):
- When \( x = 0 \):
\[ z = 2(0) - 3 = -3 \]
- When \( x = 1 \):
\[ z = 2(1) - 3 = -1 \]
So, the new limits of integration are from \( z = -3 \) to \( z = -1 \).

### Step 3: Rewrite the integral
Substituting \( z \) and \( dx \) into the integral, we have:
\[ \int_0^1 \frac{1}{2x-3} \, dx = \int_{-3}^{-1} \frac{1}{z} \cdot \frac{dz}{2} \]
This simplifies to:
\[ \frac{1}{2} \int_{-3}^{-1} \frac{1}{z} \, dz \]

### Step 4: Evaluate the integral
The integral \( \int \frac{1}{z} \, dz \) is \( \log |z| \). Thus, we have:
\[ \frac{1}{2} \left[ \log |z| \right]_{-3}^{-1} \]

### Step 5: Substitute the limits
Now we substitute the limits:
\[ = \frac{1}{2} \left( \log |-1| - \log |-3| \right) \]
Since \( \log |-1| = \log 1 = 0 \) and \( \log |-3| = \log 3 \), we get:
\[ = \frac{1}{2} \left( 0 - \log 3 \right) = -\frac{1}{2} \log 3 \]

### Final Answer
Thus, the value of the integral \( \int_0^1 \frac{1}{2x-3} \, dx \) is:
\[ -\frac{1}{2} \log 3 \]