$\int (e^{x\log a} + e^{a\log x}) dx$ is equal to (where $a > 1$)

Given: $\int (e^{x\log a} + e^{a\log x}) dx$

For the first term $e^{x\log a}$:

Using the property $n\log b = \log b^n$:

$x\log a = \log a^x$

Therefore:

$e^{x\log a} = e^{\log a^x} = a^x$

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For the second term $e^{a\log x}$:

$a\log x = \log x^a$

Therefore:

$e^{a\log x} = e^{\log x^a} = x^a$

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The integral simplifies to:

$\int (a^x + x^a) dx$

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For $\int a^x dx$:

$\int a^x dx = \frac{a^x}{\log a} + C_1$

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For $\int x^a dx$:

$\int x^a dx = \frac{x^{a+1}}{a+1} + C_2$

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Combining the results:

$\int (a^x + x^a) dx = \frac{a^x}{\log a} + \frac{x^{a+1}}{a+1} + C$

where $C$ is an arbitrary constant.