- #1
happyg1
- 308
- 0
hi,
I'm working on thiis:
[tex].25 = \int_{-\infty}^m\frac{e^{x}}{(1+e^{-x})^2}dx[/tex]
I let [tex]u=1+e^{-x} , du=-e{-x}dx[/tex]
which gives:
[tex]\frac{1}{u}=\frac{1}{1+e^{-x}}|_{-\infty}^m[/tex]
then I solve it out and I get
[tex]\frac{1}{1+e^{-m}}-1=.25[/tex]
I have tried to solve this and I keep getting a negative ln. I moved the 1 to the RHS and then took ln of both sides. I can't get it to work.
help
CC
I'm working on thiis:
[tex].25 = \int_{-\infty}^m\frac{e^{x}}{(1+e^{-x})^2}dx[/tex]
I let [tex]u=1+e^{-x} , du=-e{-x}dx[/tex]
which gives:
[tex]\frac{1}{u}=\frac{1}{1+e^{-x}}|_{-\infty}^m[/tex]
then I solve it out and I get
[tex]\frac{1}{1+e^{-m}}-1=.25[/tex]
I have tried to solve this and I keep getting a negative ln. I moved the 1 to the RHS and then took ln of both sides. I can't get it to work.
help
CC