indefinite-integration

Question: $\int_{{}}^{{}}{\frac{dx}{1+{{e}^{x}}}=}$



1) $\log (1+{{e}^{x}})$
2) $-\log (1+{{e}^{-x}})$
3) $-\log (1-{{e}^{-x}})$
4) $\log ({{e}^{-x}}+{{e}^{-2x}})$
Solution: Explanation: No Explanation
Integration by substitution

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