`inte^(x) .(a+be^(x))^(n) dx`

Answer

To solve the integral \( \int e^x (a + b e^x)^n \, dx \), we can use substitution to simplify the integration process. Here’s a step-by-step solution:

### Step 1: Substitution
Let:
\[ t = a + b e^x \]
Then, we differentiate both sides with respect to \( x \):
\[ \frac{dt}{dx} = b e^x \]
This implies:
\[ dt = b e^x \, dx \quad \Rightarrow \quad e^x \, dx = \frac{dt}{b} \]

### Step 2: Rewrite the Integral
Now we can rewrite the integral in terms of \( t \):
\[ \int e^x (a + b e^x)^n \, dx = \int (t)^n \cdot \frac{dt}{b} \]
This simplifies to:
\[ \frac{1}{b} \int t^n \, dt \]

### Step 3: Integrate
The integral of \( t^n \) is given by:
\[ \int t^n \, dt = \frac{t^{n+1}}{n+1} + C \]
Thus, we have:
\[ \frac{1}{b} \int t^n \, dt = \frac{1}{b} \left( \frac{t^{n+1}}{n+1} + C \right) = \frac{t^{n+1}}{b(n+1)} + \frac{C}{b} \]

### Step 4: Substitute Back
Now we substitute back \( t = a + b e^x \):
\[ \frac{(a + b e^x)^{n+1}}{b(n+1)} + C \]

### Final Answer
Thus, the final answer for the integral is:
\[ \int e^x (a + b e^x)^n \, dx = \frac{(a + b e^x)^{n+1}}{b(n+1)} + C \]

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