Evaluate the integerals.
`int (1)/(7x+3) dx on I sub R \\{-(3)/(7)}.`

Answer

To evaluate the integral \( \int \frac{1}{7x + 3} \, dx \), we can follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ \int \frac{1}{7x + 3} \, dx \]
To simplify the integration, we can multiply and divide the integrand by 7:
\[ \int \frac{1}{7} \cdot \frac{7}{7x + 3} \, dx \]
This allows us to factor out \( \frac{1}{7} \):
\[ \frac{1}{7} \int \frac{7}{7x + 3} \, dx \]

### Step 2: Substitution
Next, we will use substitution. Let:
\[ d = 7x + 3 \]
Then, differentiating both sides with respect to \( x \), we get:
\[ \frac{dt}{dx} = 7 \quad \Rightarrow \quad dt = 7 \, dx \quad \Rightarrow \quad dx = \frac{dt}{7} \]
Now we can substitute \( d \) and \( dx \) into the integral:
\[ \frac{1}{7} \int \frac{7}{d} \cdot \frac{dt}{7} \]
This simplifies to:
\[ \frac{1}{7} \int \frac{1}{d} \, dt \]

### Step 3: Integrate
Now we can integrate:
\[ \frac{1}{7} \ln |d| + C \]
Substituting back for \( d \):
\[ \frac{1}{7} \ln |7x + 3| + C \]

### Final Answer
Thus, the final result of the integral is:
\[ \int \frac{1}{7x + 3} \, dx = \frac{1}{7} \ln |7x + 3| + C \]