Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SIGN UP FOR FREE
Hi there, use the substitution: u = e^x du = e^x * dx
Sorry I will be typing in short posts, because it is easier for me to see what I'm doing
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

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

You will get: \[(1/e ^{x}) \int\limits_{}^{} e ^{x} / ( 1 + e ^{x})\] Now you can replace x and dx by u and du
You will get: \[(1/e ^{x}) \int\limits_{}^{} e ^{x} / ( 1 + e ^{x})\]dx Now you can replace x and dx by u and
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
\[1/ e ^{x} \int\limits_{}^{} e^x * dx/ ( 1 + e^x) = 1 / e^x \int\limits_{}^{} du / (1+u)\]
Now remember we let u = e^x, so plug that into the 1/e^x and move everything into the integral
\[\int\limits_{}^{}du / [u(1+u)]\]
Now use partial fractions: ( A / u ) + ( B / 1 + u ) = 1 / [u(1+u)]
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
Rewrite your integral as: \[\int\limits\limits_{}^{}du / [u(1+u)]=\int\limits_{}^{}(1/u)du + \int\limits_{}^{}( -1*du) / (1+u)\]
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!
x - ln (1+e^x) +C
you there?
yeah tryin to figure out ur question

Not the answer you are looking for?

Search for more explanations.

Ask your own question