Question

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

Answer 1

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