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