Evaluate $\displaystyle \int \frac{1}{e^x + 1} dx$.
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$\displaystyle -ln(e^x+1)+x$
$\displaystyle \frac{1}{e^x + 1} \times \frac{e^{-x}}{e^{-x}}= \frac{e^{-x}}{1 + e^{-x}}$ Can you go from here?
Originally Posted by icemanfan Evaluate $\displaystyle \int \frac{1}{e^x + 1} dx$. $\displaystyle = \int \frac{e^x + 1 - e^x}{e^x + 1} dx$ $\displaystyle = \int \frac{e^x + 1}{e^x + 1}dx - \int \frac{e^x}{e^x + 1} dx$ $\displaystyle = x - \ln(e^x + 1) + c$
$\displaystyle \int \frac{1}{1 + e^x}dx = \int \frac{e^{-x}}{1 + e^{-x}} dx = -\ln(1 + e^{-x}) + C$ Correct?
Originally Posted by icemanfan $\displaystyle \int \frac{1}{1 + e^x}dx = \int \frac{e^{-x}}{1 + e^{-x}} dx = -\ln(1 + e^{-x}) + C$ Correct? You lost an x and your e should be to the positive power.
Originally Posted by dwsmith You lost an x and your e should be to the positive power. Nah it's right I think. Someone show me how the two answers we have given are equiv though..?
Last edited by Deadstar; Apr 12th 2010 at 05:29 PM.
Originally Posted by Deadstar Nah it's right I think. Someone show me how the two answers we have given are equiv though..? Yeah they are equivalent. I checked them in Maple; however, I have no idea how to show that they are.
Originally Posted by Deadstar Nah it's right I think. Someone show me how the two answers we have given are equiv though..? -ln(1-e^-x) = -ln(1 - 1/e^x) = -ln[(e^x - 1)/e^x] = -[ln(e^x - 1) - lne^x] = x- ln(e^x - 1)
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