Question

$\int \frac{e^{x-1}+x^{e-1}}{e^x+x^e}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{e^{x-1}+x^{e-1}}{e^x+x^e}dx \\ \mathrm{Putting}{e}^x+x^e=t \\ \Rightarrow e^x+e{x}^{e-1}=\frac{dt}{dx} \\ \Rightarrow e\left(e^{x-1}+x^{e-1}\right)=\frac{dt}{dx} \\ \Rightarrow \left(e^{x-1}+x^{e-1}\right)dx=\frac{dt}{e} \\ \therefore I=\frac{1}{e}\int \frac{1}{t}dt \\ =\frac{1}{e}\ln \left|t\right|+\mathrm{C} \\ =\frac{1}{e}\ln \left|e^x+x^e\right|+\mathrm{C}\left[\because \mathit{}t={\mathrm{e}}^{\mathit{x}}+x^{\mathrm{e}}\right] \end{gathered}$$