What is the value of `inta^(x)e^(x)dx` ?

Options

A. `(a^(x)e^(x))/(lna)+C`
B. `a^(x)e^(x)+C`
C. `(a^(x)e^(x))/(ln(ae))+C`
D. None of these

Answer

To solve the integral \( \int a^x e^x \, dx \), we will use the method of integration by parts. Here’s a step-by-step solution:

### Step 1: Identify \( u \) and \( dv \)
We will choose:
- \( u = a^x \) (which we will differentiate)
- \( dv = e^x \, dx \) (which we will integrate)

### Step 2: Differentiate \( u \) and Integrate \( dv \)
Now we find \( du \) and \( v \):
- Differentiate \( u \):
\[ du = a^x \ln(a) \, dx \]
- Integrate \( dv \):
\[ v = e^x \]

### Step 3: Apply the Integration by Parts Formula
The integration by parts formula is given by:
\[ \int u \, dv = uv - \int v \, du \]
Substituting our values into this formula:
\[ \int a^x e^x \, dx = a^x e^x - \int e^x (a^x \ln(a)) \, dx \]

### Step 4: Simplify the Integral
Now, we can factor out \( \ln(a) \):
\[ \int a^x e^x \, dx = a^x e^x - \ln(a) \int a^x e^x \, dx \]

### Step 5: Solve for the Integral
Let \( I = \int a^x e^x \, dx \). Then we have:
\[ I = a^x e^x - \ln(a) I \]
Rearranging gives:
\[ I + \ln(a) I = a^x e^x \]
\[ I(1 + \ln(a)) = a^x e^x \]
Thus, we can solve for \( I \):
\[ I = \frac{a^x e^x}{1 + \ln(a)} + C \]

### Final Answer
Therefore, the value of the integral is:
\[ \int a^x e^x \, dx = \frac{a^x e^x}{1 + \ln(a)} + C \]