`int1/(a+be^(x))dx=`

Options

A. `1/blogabs(a+be^(x))+c`
B. `-1/blogabs(a+be^(x))+c`
C. `-1/blogabs(ae^(x)+1)+c`
D. `1/alogabs(e^(x)/(a+be^(x)))+c`

Answer

To solve the integral \( \int \frac{1}{a + b e^x} \, dx \), we can follow these steps:

### Step 1: Rewrite the Integral
Let \( I = \int \frac{1}{a + b e^x} \, dx \).

### Step 2: Factor out \( e^x \)
We can rewrite the integrand by factoring out \( e^x \):
\[ I = \int \frac{e^{-x}}{e^{-x}(a + b e^x)} \, dx = \int \frac{e^{-x}}{a e^{-x} + b} \, dx \]

### Step 3: Substitution
Now, let \( t = a + b e^x \). Then, we differentiate:
\[ dt = b e^x \, dx \quad \Rightarrow \quad dx = \frac{dt}{b e^x} = \frac{dt}{b \left( \frac{t - a}{b} \right)} = \frac{dt}{t - a} \]

### Step 4: Substitute in the Integral
Substituting \( t \) into the integral gives:
\[ I = \int \frac{e^{-x}}{t} \cdot \frac{dt}{b \left( \frac{t - a}{b} \right)} = \int \frac{e^{-x}}{t} \cdot \frac{dt}{t - a} \]

### Step 5: Express \( e^{-x} \)
From our substitution \( t = a + b e^x \), we can express \( e^{-x} \) as:
\[ e^{-x} = \frac{t - a}{b} \]
Thus, the integral becomes:
\[ I = \int \frac{(t - a)/b}{t} \cdot \frac{dt}{b} = \frac{1}{b^2} \int \left( 1 - \frac{a}{t} \right) dt \]

### Step 6: Integrate
Now, we can integrate:
\[ I = \frac{1}{b^2} \left( t - a \ln |t| \right) + C \]

### Step 7: Substitute Back for \( t \)
Substituting back for \( t \):
\[ I = \frac{1}{b^2} \left( a + b e^x - a \ln |a + b e^x| \right) + C \]

### Final Answer
Thus, the final result is:
\[ \int \frac{1}{a + b e^x} \, dx = \frac{1}{b} \ln |a + b e^x| + C \]