Integrate the following functions with respect to `t`

Answer

To solve the problem of integrating the given functions with respect to \( t \), we will go through each function step by step.

### Problem 1: Integrate \( 3t^2 - 2t \)

1. **Separate the terms**:
\[ \int (3t^2 - 2t) \, dt = \int 3t^2 \, dt - \int 2t \, dt \]

2. **Integrate each term**:
- For \( \int 3t^2 \, dt \):
\[ = 3 \cdot \frac{t^{2+1}}{2+1} = 3 \cdot \frac{t^3}{3} = t^3 \]
- For \( \int 2t \, dt \):
\[ = 2 \cdot \frac{t^{1+1}}{1+1} = 2 \cdot \frac{t^2}{2} = t^2 \]

3. **Combine the results**:
\[ \int (3t^2 - 2t) \, dt = t^3 - t^2 + C \]

### Problem 2: Integrate \( 4 \cos t + t^2 \)

1. **Separate the terms**:
\[ \int (4 \cos t + t^2) \, dt = \int 4 \cos t \, dt + \int t^2 \, dt \]

2. **Integrate each term**:
- For \( \int 4 \cos t \, dt \):
\[ = 4 \sin t \]
- For \( \int t^2 \, dt \):
\[ = \frac{t^{2+1}}{2+1} = \frac{t^3}{3} \]

3. **Combine the results**:
\[ \int (4 \cos t + t^2) \, dt = 4 \sin t + \frac{t^3}{3} + C \]

### Problem 3: Integrate \( (2t - 4)^{-4} \)

1. **Substitution**: Let \( x = 2t - 4 \). Then, \( dx = 2 \, dt \) or \( dt = \frac{dx}{2} \).

2. **Rewrite the integral**:
\[ \int (2t - 4)^{-4} \, dt = \int x^{-4} \cdot \frac{dx}{2} = \frac{1}{2} \int x^{-4} \, dx \]

3. **Integrate**:
\[ = \frac{1}{2} \cdot \frac{x^{-4+1}}{-4+1} = \frac{1}{2} \cdot \frac{x^{-3}}{-3} = -\frac{1}{6} x^{-3} \]

4. **Substitute back**:
\[ = -\frac{1}{6} (2t - 4)^{-3} + C \]

### Problem 4: Integrate \( \frac{1}{6t - 1} \)

1. **Substitution**: Let \( x = 6t - 1 \). Then, \( dx = 6 \, dt \) or \( dt = \frac{dx}{6} \).

2. **Rewrite the integral**:
\[ \int \frac{1}{6t - 1} \, dt = \int \frac{1}{x} \cdot \frac{dx}{6} = \frac{1}{6} \int \frac{1}{x} \, dx \]

3. **Integrate**:
\[ = \frac{1}{6} \ln |x| + C \]

4. **Substitute back**:
\[ = \frac{1}{6} \ln |6t - 1| + C \]

### Final Answers:
1. \( \int (3t^2 - 2t) \, dt = t^3 - t^2 + C \)
2. \( \int (4 \cos t + t^2) \, dt = 4 \sin t + \frac{t^3}{3} + C \)
3. \( \int (2t - 4)^{-4} \, dt = -\frac{1}{6} (2t - 4)^{-3} + C \)
4. \( \int \frac{1}{6t - 1} \, dt = \frac{1}{6} \ln |6t - 1| + C \)