`int (e^(xloga)+e^(alogx))dx`

Answer

To solve the integral \( I = \int \left( e^{x \log a} + e^{a \log x} \right) dx \), we can follow these steps:

### Step 1: Simplify the terms using properties of exponents and logarithms
We start with the expression:
\[ I = \int \left( e^{x \log a} + e^{a \log x} \right) dx \]
Using the property \( e^{\log b} = b \), we can rewrite the terms:
\[ e^{x \log a} = a^x \quad \text{and} \quad e^{a \log x} = x^a \]
Thus, we can rewrite the integral as:
\[ I = \int \left( a^x + x^a \right) dx \]

### Step 2: Integrate each term separately
Now we can split the integral:
\[ I = \int a^x \, dx + \int x^a \, dx \]
We know the integral of \( a^x \) and \( x^a \):
1. The integral of \( a^x \) is:
\[ \int a^x \, dx = \frac{a^x}{\ln a} + C_1 \]
2. The integral of \( x^a \) is:
\[ \int x^a \, dx = \frac{x^{a+1}}{a+1} + C_2 \]

### Step 3: Combine the results
Putting it all together, we have:
\[ I = \frac{a^x}{\ln a} + \frac{x^{a+1}}{a+1} + C \]
where \( C = C_1 + C_2 \) is the constant of integration.

### Final Answer
Thus, the final result for the integral is:
\[ I = \frac{a^x}{\ln a} + \frac{x^{a+1}}{a+1} + C \]