Integrate the following `:` `int(dt)/((6t-1))`

Explanation

To solve the integral \( \int \frac{dt}{6t - 1} \), we can follow these steps:

### Step 1: Identify the integral
We start with the integral:
\[ I = \int \frac{dt}{6t - 1} \]

### Step 2: Use substitution
To simplify the integral, we can use a substitution. Let:
\[ y = 6t - 1 \]
Then, we differentiate \( y \) with respect to \( t \):
\[ dy = 6 \, dt \quad \Rightarrow \quad dt = \frac{dy}{6} \]

### Step 3: Substitute into the integral
Now we substitute \( y \) and \( dt \) into the integral:
\[ I = \int \frac{dt}{6t - 1} = \int \frac{\frac{dy}{6}}{y} \]
This simplifies to:
\[ I = \frac{1}{6} \int \frac{dy}{y} \]

### Step 4: Integrate
The integral \( \int \frac{dy}{y} \) is a standard integral, which results in:
\[ \int \frac{dy}{y} = \ln |y| + C \]
Thus, we have:
\[ I = \frac{1}{6} \ln |y| + C \]

### Step 5: Substitute back for \( y \)
Now we substitute back \( y = 6t - 1 \):
\[ I = \frac{1}{6} \ln |6t - 1| + C \]

### Final Answer
Therefore, the final result of the integral is:
\[ \int \frac{dt}{6t - 1} = \frac{1}{6} \ln |6t - 1| + C \]

---