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How to integrate e^(-2x)/(1+e^(-x)) dx?
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\[\int\limits_{}^{}\frac{ e ^{-2x} }{ 1+e ^{-x} } dx\]
@ganeshie8 @jabez177 @inkyvoyd @Conqueror @mathmath333
i'd go \(u = e^{-x}, du = - e^{-x} dx = -u ~dx \) that already looks promising so \(\int ~ \dfrac{u^2}{1 + u } . \dfrac{-1}{u} ~ du\) \(= - \int ~ \dfrac{u}{1 + u } ~ du\) \(=- \int ~ \dfrac{1 + u - 1}{1 + u } ~ du\) \(=- \int ~ 1 - \dfrac{ 1}{1 + u } ~ du\) bish bosh
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