(i) `int_0^1 (dx)/(3x-4)` (ii) `int_1^3 (dx)/(7-2x)`

Answer

Let's solve the integrals step by step.

### Part (i): Evaluate \( I_1 = \int_0^1 \frac{dx}{3x - 4} \)

1. **Substitution**: Let \( t = 3x - 4 \). Then, differentiate both sides:
\[ dt = 3dx \quad \Rightarrow \quad dx = \frac{dt}{3} \]

2. **Change the limits**: When \( x = 0 \):
\[ t = 3(0) - 4 = -4 \]
When \( x = 1 \):
\[ t = 3(1) - 4 = -1 \]
So the new limits for \( t \) are from \(-4\) to \(-1\).

3. **Substituting in the integral**:
\[ I_1 = \int_{-4}^{-1} \frac{1}{t} \cdot \frac{dt}{3} = \frac{1}{3} \int_{-4}^{-1} \frac{dt}{t} \]

4. **Integrate**:
\[ I_1 = \frac{1}{3} \left[ \ln |t| \right]_{-4}^{-1} = \frac{1}{3} \left( \ln |-1| - \ln |-4| \right) \]
Since \( \ln |-1| = 0 \):
\[ I_1 = \frac{1}{3} (0 - \ln 4) = -\frac{1}{3} \ln 4 \]

### Part (ii): Evaluate \( I_2 = \int_1^3 \frac{dx}{7 - 2x} \)

1. **Substitution**: Let \( t = 7 - 2x \). Then, differentiate:
\[ dt = -2dx \quad \Rightarrow \quad dx = -\frac{dt}{2} \]

2. **Change the limits**: When \( x = 1 \):
\[ t = 7 - 2(1) = 5 \]
When \( x = 3 \):
\[ t = 7 - 2(3) = 1 \]
So the new limits for \( t \) are from \( 5 \) to \( 1 \).

3. **Substituting in the integral**:
\[ I_2 = \int_{5}^{1} \frac{1}{t} \left(-\frac{dt}{2}\right) = -\frac{1}{2} \int_{5}^{1} \frac{dt}{t} \]

4. **Integrate**:
\[ I_2 = -\frac{1}{2} \left[ \ln |t| \right]_{5}^{1} = -\frac{1}{2} \left( \ln |1| - \ln |5| \right) \]
Since \( \ln |1| = 0 \):
\[ I_2 = -\frac{1}{2} (0 - \ln 5) = \frac{1}{2} \ln 5 \]

### Final Answers:
- \( I_1 = -\frac{1}{3} \ln 4 \)
- \( I_2 = \frac{1}{2} \ln 5 \)