Evaluate the following integration
`int 2^(x)*e^(x)*dx`

Answer

To evaluate the integral \( I = \int 2^x e^x \, dx \), we will use the method of integration by parts.

### Step-by-Step Solution:

1. **Choose \( u \) and \( dv \)**:
Let:
\[ u = 2^x \quad \text{and} \quad dv = e^x \, dx \]

2. **Differentiate \( u \) and Integrate \( dv \)**:
Now, we need to find \( du \) and \( v \):
\[ du = \frac{d}{dx}(2^x) \, dx = 2^x \log(2) \, dx \]
\[ v = \int e^x \, dx = e^x \]

3. **Apply the Integration by Parts Formula**:
The integration by parts formula is:
\[ \int u \, dv = uv - \int v \, du \]
Substituting the values we have:
\[ I = 2^x e^x - \int e^x (2^x \log(2)) \, dx \]

4. **Simplify the Integral**:
The integral we have now is:
\[ I = 2^x e^x - \log(2) \int 2^x e^x \, dx \]
Notice that the integral on the right side is the same as \( I \). Thus, we can rewrite it as:
\[ I = 2^x e^x - \log(2) I \]

5. **Combine Like Terms**:
Rearranging the equation gives:
\[ I + \log(2) I = 2^x e^x \]
\[ I(1 + \log(2)) = 2^x e^x \]

6. **Solve for \( I \)**:
Finally, we can solve for \( I \):
\[ I = \frac{2^x e^x}{1 + \log(2)} + C \]
where \( C \) is the constant of integration.

### Final Answer:
\[ \int 2^x e^x \, dx = \frac{2^x e^x}{1 + \log(2)} + C \]