Tricky integral

  • Thread starter fk378
  • Start date
  • #1
fk378
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Homework Statement


Integrate.
1/ [e^x (1-e^(-2x))^1/2]


Homework Equations


integration by parts


The Attempt at a Solution


First I split up the integral so it would be (1/e^-x)[(1/(1-e^-2x)^1/2]
Then I set u=e^-x, dv= (1-e^-2x)^-1/2
du= -e^-x
For my v I got 2(1-e^-2x)^1/2 but I don't know how to go about with the e^-2x, as in how to finish solving for the v-part.

I tried u-substitution and I get another term in the denominator. Then when I tried to differentiate the answer I originally got for v, I did not have anything in the denominator... can anyone help?
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
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[tex]\int \frac{1}{e^x \sqrt{1-e^{-2x}}} dx[/tex]



try [itex]t^2=1-e^{-2x}[/itex]
 
  • #3
fk378
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Is that valid? I just tried to solve it using your suggestion and with u=e^-x. With the u-subst I end up getting the arcsin integral, but I get something different with the t^2 substitution.
 
  • #4
rocomath
1,755
1
That's what I did too and verified it by taking it's derivative, but Mathematica gives a totally different solution.

I changed mine to:

[tex]\int\frac{e^{-x}}{\sqrt{1-(e^{-x})^2}}dx[/tex]

Used:

[tex]u=e^{-x} \rightarrow \ln{u}=-x[/tex]
 
  • #5
fk378
367
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Oh interesting, why did you take the natural log? Where does that come into play?
 
  • #6
rocomath
1,755
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I let:

[tex]u=e^{-x}[/tex] be my u-sub

[tex]-\frac{du}{u}=dx[/tex] be my du-sub
 

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