I honestly have no idea... how do i integrate 2/ ((e^-x)+1)??
X XowakkadoodleoX New member Joined Mar 27, 2012 Messages 2 Mar 27, 2012 #1 I honestly have no idea... how do i integrate 2/ ((e^-x)+1)??
D Deleted member 4993 Guest Mar 27, 2012 #2 XowakkadoodleoX said: I honestly have no idea... how do i integrate 2/ ((e^-x)+1)?? Click to expand... \(\displaystyle \int\dfrac{1}{e^{-x}+1}dx \ = \ \int\dfrac{e^x}{e^{x}+1}dx\) and continue....
XowakkadoodleoX said: I honestly have no idea... how do i integrate 2/ ((e^-x)+1)?? Click to expand... \(\displaystyle \int\dfrac{1}{e^{-x}+1}dx \ = \ \int\dfrac{e^x}{e^{x}+1}dx\) and continue....
H HallsofIvy Elite Member Joined Jan 27, 2012 Messages 7,763 Mar 27, 2012 #3 Equivalently, let \(\displaystyle u= e^{-x}+ 1\) so that \(\displaystyle du= -e^{-x}dx\), \(\displaystyle -du/e^{-x}= -du/(e^{-x}+1)- 1)= -du/(u-1)= dx\). Last edited: Mar 27, 2012
Equivalently, let \(\displaystyle u= e^{-x}+ 1\) so that \(\displaystyle du= -e^{-x}dx\), \(\displaystyle -du/e^{-x}= -du/(e^{-x}+1)- 1)= -du/(u-1)= dx\).
