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प्रश्न
उत्तर
\[Let\ e^x = t . Then, e^x dx = dt\]
\[When\ x = 0, t = 1\ and\ x = 1, t = e\]
\[ \therefore I = \int_0^1 \frac{e^x}{1 + e^{2x}} d x\]
\[ \Rightarrow I = \int_1^e \frac{dt}{1 + t^2}\]
\[ \Rightarrow I = \left[ \tan^{- 1} x \right]_1^e \]
\[ \Rightarrow I = \tan^{- 1} e - \tan^{- 1} 1\]
\[ \Rightarrow I = \tan^{- 1} e - \frac{\pi}{4}\]
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