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$\int \frac{e^{x}}{(1 + e^{2 x}) \tan^{- 1} (e^{x})} d x =$

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$\log | \tan^{- 1} (e^{x}) | + c$

$\log | \cot^{- 1} (e^{x}) | + c$

$\log | 1 + e^{2 x} | + c$

$\log | 1 + e^{x} | + c$

answer is A.

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Detailed Solution

$\int \frac{e^{x}}{(1 + e^{2 x}) \tan^{- 1} (e^{x})} d x$

$p u t \tan^{- 1} e^{x} = t$

$= \int \frac{d t}{t} \frac{1}{1 + {(e^{x})}^{2}} . e^{x} d x = d t$

$= \log | t | + c$

$= \log | \tan^{- 1} e^{x} | + c$

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