$\int \frac{e^{x-1}+x^{e-1}}{e^x+x} d x$ is equal to
Answer & explanation
Correct answer: option 1
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.