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OpenStudy (anonymous):

integral 1 / ( 1 + e^x) dx .

OpenStudy (bahrom7893):

Hi there, use the substitution: u = e^x du = e^x * dx

OpenStudy (bahrom7893):

Sorry I will be typing in short posts, because it is easier for me to see what I'm doing

OpenStudy (bahrom7893):

Now if you let u = e^x and du = e^x * dx, then you have Integral ( 1 / ( 1 + u ) * dx ). NOTE you are missing the e^x in the dx, so multiply and divide the integral by e^x

OpenStudy (bahrom7893):

You will get: \[(1/e ^{x}) \int\limits_{}^{} e ^{x} / ( 1 + e ^{x})\] Now you can replace x and dx by u and du

OpenStudy (bahrom7893):

You will get: \[(1/e ^{x}) \int\limits_{}^{} e ^{x} / ( 1 + e ^{x})\]dx Now you can replace x and dx by u and

OpenStudy (bahrom7893):

Sorry I forgot the dx in the previous part. So now you have e^x * dx = du on top and 1 + u on the bottom

OpenStudy (bahrom7893):

\[1/ e ^{x} \int\limits_{}^{} e^x * dx/ ( 1 + e^x) = 1 / e^x \int\limits_{}^{} du / (1+u)\]

OpenStudy (bahrom7893):

Now remember we let u = e^x, so plug that into the 1/e^x and move everything into the integral

OpenStudy (bahrom7893):

\[\int\limits_{}^{}du / [u(1+u)]\]

OpenStudy (bahrom7893):

Now use partial fractions: ( A / u ) + ( B / 1 + u ) = 1 / [u(1+u)]

OpenStudy (bahrom7893):

multiply everything by u * ( 1 + u ) so that you get: A * (1+u) + B * u = 1 A + Au + Bu = 1 + 0 A + u(A+B) = 1 + 0 A = 1; A+B = 0 => A = 1; B = -1

OpenStudy (bahrom7893):

Rewrite your integral as: \[\int\limits\limits_{}^{}du / [u(1+u)]=\int\limits_{}^{}(1/u)du + \int\limits_{}^{}( -1*du) / (1+u)\]

OpenStudy (bahrom7893):

The first part is just ln|u| and for the second part of the integral, let a = 1+u, da=du; so the answer to the integral so far is: Ln|u| - Ln|1+u|; u = e^x => Answer is Ln(e^x) - Ln(1+e^x) = x - ln (1+e^x) +C That's the final answer!

OpenStudy (bahrom7893):

x - ln (1+e^x) +C

OpenStudy (anonymous):

you there?

OpenStudy (bahrom7893):

yeah tryin to figure out ur question

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