$\int \frac{e^{x}}{1 + e^{2 x}} d x$

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Solution

Let $e^{x} = t \Rightarrow e^{x} d x = d t$

So, $\int \frac{e^{x} d x}{1 + e^{2 x}} = \int \frac{d t}{1 + t^{2}} = {tan}^{- 1} t + c = {tan}^{- 1} e^{x} + c$

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C

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