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∫dx /√[1+e^(2x)]
let
e^x = tanu
e^x dx = (secu)^2 du
dx = [(secu)^2 / tanu ] du
∫dx /√[1+e^(2x)]
=∫[(secu)^2 / tanu ] du / (secu)
=∫ (secu/ tanu) du
=∫ cscu du
=ln|cscu - cotu |+ C
=ln|√[1+e^(2x)] /e^x - 1/e^x |+ C
=ln|√[1+e^(2x)] -1 | -x+ C
let
e^x = tanu
e^x dx = (secu)^2 du
dx = [(secu)^2 / tanu ] du
∫dx /√[1+e^(2x)]
=∫[(secu)^2 / tanu ] du / (secu)
=∫ (secu/ tanu) du
=∫ cscu du
=ln|cscu - cotu |+ C
=ln|√[1+e^(2x)] /e^x - 1/e^x |+ C
=ln|√[1+e^(2x)] -1 | -x+ C
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