$\int e^{(x \log 5)}e^x dx$, is:

Where $C$ is the constant of integration.

$e^{x \log 5} = e^{\log 5^x} = 5^x$

The integral becomes:

$\int 5^x \cdot e^x \, dx$

When multiplying exponentials with the same power:

$5^x \cdot e^x = (5 \times e)^x = (5e)^x$

The integral simplifies to:

$\int (5e)^x \, dx$

Using the formula $\int a^x \, dx = \frac{a^x}{\log a} + C$ where $\log$ denotes the natural logarithm:

$\int (5e)^x \, dx = \frac{(5e)^x}{\log(5e)} + C$

Therefore, the answer is:

$\dfrac{(5e)^x}{\log(5e)} + C$