$\int \frac{e^{x-1}+x^{e-1}}{e^x+x} d x$ is equal to

$\frac{1}{e} \ln \left(x^e+e^x\right)+c$

Let $I=\int \frac{e^{x-1}+x^{e-1}}{e^x+x^e} d x$

Let $e^{x}+x^{e}=t \Rightarrow\left(e^{x}+e . x^{e-1}\right) dx=dt$

$\Rightarrow e\left(e^{x-1}+x^{e-1}\right) dx=dt$

∴  $I=\frac{1}{e} \int \frac{d t}{t}=\frac{1}{e} \ln \left(e^{x}+x^{e}\right)+c$

Hence (1) is the correct answer.