Integration problem

  • Thread starter cloveryeah
  • Start date
  • #1
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Homework Statement


integrate 1/(1+e^x) dx

Homework Equations




The Attempt at a Solution


firstly i let t=1+e^x
and then i come to : integrate 1/(t^2-1)
and then i put t=secx
.
.
.
but then the final ans is -1/2 ln | 2/e^x +1 |

it should be 1 instead of 2, i hv checked for the steps for so many times, but found nothing wrong
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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You have done the substitution wrong.

If [itex]t= 1+ e^x[/itex] then [itex]dt= e^xdx= (t- 1)dx[/itex] so that [itex]\frac{1}{t- 1}= dx[/itex].

[tex]\int \frac{1}{1+ e^x}dx= \int \frac{1}{t} \frac{dt}{t- 1}= \int \frac{1}{t(t- 1)} dt[/tex]

NOT [itex]\int \frac{1}{t^2- 1} dt[/itex]
 
  • #3
STEMucator
Homework Helper
2,075
140
Perhaps you should multiply the top and bottom of the expression by ##e^{-x}## and see what happens when you substitute ##u = e^{-x}##.
 
  • #4
You have done the substitution wrong.

If [itex]t= 1+ e^x[/itex] then [itex]dt= e^xdx= (t- 1)dx[/itex] so that [itex]\frac{1}{t- 1}= dx[/itex].

[tex]\int \frac{1}{1+ e^x}dx= \int \frac{1}{t} \frac{dt}{t- 1}= \int \frac{1}{t(t- 1)} dt[/tex]

NOT [itex]\int \frac{1}{t^2- 1} dt[/itex]
then you can use partial fraction i.e. create (t)- (t-1) in the numerator
 

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