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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

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# Integral calculation help

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