Evaluate the following integrals:

Multiplying and dividing the numerator by e we get the given as

$\Rightarrow \frac{1}{e} \int \frac{e^{x}+e x^{e-1}}{e^{x}+x^{e}} d x \ldots$ (1)

Assume $e^{x}+x^{e}=t$

$\Rightarrow d\left(e^{x}+x^{e}\right)=d t$

$\Rightarrow e^{x}+e x^{e-1}=d t$

Substituting $\mathrm{t}$ and dt in equation 1 we get

$\Rightarrow \frac{1}{\mathrm{e}} \int \frac{\mathrm{d} t}{t}$

$=\ln |t|+c$

But $t=e^{x}+x^{e}$

$\therefore \ln \left|e^{x}+x^{e}\right|+c$