Question #250743
  1. Intrigration of dx/1+e^2x
1
Expert's answer
2021-10-14T04:27:44-0400
1e2x+1dx=e2x1+e2xdx\int\dfrac{1}{e^{2x}+1}dx=\int\dfrac{e^{-2x}}{1+e^{-2x}}dx

u=1+e2x,du=2e2xdxu=1+e^{-2x}, du=-2e^{-2x}dx

1e2x+1dx=e2x1+e2xdx=121udu\int\dfrac{1}{e^{2x}+1}dx=\int\dfrac{e^{-2x}}{1+e^{-2x}}dx=-\dfrac{1}{2}\int\dfrac{1}{u}du

=12ln(u)+C=12ln(1+e2x)+C=-\dfrac{1}{2}\ln(|u|)+C=-\dfrac{1}{2}\ln(1+e^{-2x})+C

1e2x+1dx=12ln(1+e2x)+C\int\dfrac{1}{e^{2x}+1}dx=-\dfrac{1}{2}\ln(1+e^{-2x})+C

=x12ln(1+e2x)+C=x-\dfrac{1}{2}\ln(1+e^{2x})+C


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS