`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

\[ I = \int_0^L \frac{dx}{ax + b} \]

we will follow these steps:

### Step 1: Substitution
Let \( t = ax + b \). Then, differentiating both sides gives:

\[ dt = a \, dx \quad \Rightarrow \quad dx = \frac{dt}{a} \]

### Step 2: Change the Limits of Integration
Now we need to change the limits of integration according to our substitution.

- When \( x = 0 \):
\[ t = a(0) + b = b \]

- When \( x = L \):
\[ t = aL + b \]

Thus, the new limits for \( t \) are from \( b \) to \( aL + b \).

### Step 3: Rewrite the Integral
Now we can rewrite the integral in terms of \( t \):

\[ I = \int_b^{aL + b} \frac{1}{t} \cdot \frac{dt}{a} = \frac{1}{a} \int_b^{aL + b} \frac{1}{t} \, dt \]

### Step 4: Integrate
The integral of \( \frac{1}{t} \) is \( \ln |t| \). Therefore, we have:

\[ I = \frac{1}{a} \left[ \ln |t| \right]_b^{aL + b} \]

### Step 5: Evaluate the Integral
Now we evaluate the definite integral:

\[ I = \frac{1}{a} \left( \ln |aL + b| - \ln |b| \right) \]

Using the property of logarithms, this can be simplified to:

\[ I = \frac{1}{a} \ln \left( \frac{aL + b}{b} \right) \]

### Final Result
Thus, the final result of the integral is:

\[ I = \frac{1}{a} \ln \left( \frac{aL + b}{b} \right) \]