`int_0^L (dx)/(ax + b)` =

Options

A. `1/b In ((aL + b)/b)`
B. `-1/a In ((aL + b)/b)`
C. `1/a In ((aL + b)/b)`
D. `-1/b In ((aL + b)/b)`

Answer

To solve the integral \( \int_0^L \frac{dx}{ax + b} \), we can follow these steps:

### Step 1: Identify the Integral
We need to evaluate the integral:
\[ \int_0^L \frac{dx}{ax + b} \]

### Step 2: Use the Formula for Integration
We know that the integral of \( \frac{1}{x} \) is \( \ln |x| + C \). For our case, we can use the substitution \( u = ax + b \). The differential \( du = a \, dx \) or \( dx = \frac{du}{a} \).

### Step 3: Change the Limits of Integration
When \( x = 0 \):
\[ u = a(0) + b = b \]
When \( x = L \):
\[ u = aL + b \]

### Step 4: Rewrite the Integral
Now we can rewrite the integral in terms of \( u \):
\[ \int_b^{aL + b} \frac{1}{u} \cdot \frac{du}{a} = \frac{1}{a} \int_b^{aL + b} \frac{du}{u} \]

### Step 5: Evaluate the Integral
The integral \( \int \frac{du}{u} \) evaluates to \( \ln |u| \):
\[ \frac{1}{a} \left[ \ln |u| \right]_b^{aL + b} = \frac{1}{a} \left( \ln |aL + b| - \ln |b| \right) \]

### Step 6: Apply the Logarithmic Property
Using the property of logarithms \( \ln A - \ln B = \ln \frac{A}{B} \):
\[ \frac{1}{a} \left( \ln |aL + b| - \ln |b| \right) = \frac{1}{a} \ln \left( \frac{aL + b}{b} \right) \]

### Final Result
Thus, the final result of the integral is:
\[ \int_0^L \frac{dx}{ax + b} = \frac{1}{a} \ln \left( \frac{aL + b}{b} \right) \]