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How do you tackle this integral

  • Thread starter Gauss M.D.
  • Start date
  • #1
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Homework Statement



Find F(x)

f(x) = 1/(1+x2)2)

Homework Equations





The Attempt at a Solution



It's actually a subproblem of another huge annoying surface integral. I tried u-sub but that landed me nowhere and partial integration led me nowhere aswell... Any pointers?
 

Answers and Replies

  • #2
Zondrina
Homework Helper
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Homework Statement



Find F(x)

f(x) = 1/(1+x2)2)

Homework Equations





The Attempt at a Solution



It's actually a subproblem of another huge annoying surface integral. I tried u-sub but that landed me nowhere and partial integration led me nowhere aswell... Any pointers?
Perfect candidate for a trig sub. Use ##x = tan(θ)## so that ##dx = sec^2(θ)dθ##.
 
  • #3
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well, 1/1+X^2 is the derivative of arctan i believe. i think there has to be a relation. u use integration by parts left u nowhere?
 
  • #4
153
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Zondrina: Aha, thanks. Obviously it just so happens that in the problem, x had an upper bound of 1. Would we have been out of luck if that hadn't been the case?
 
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  • #5
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Aha, thanks. Obviously it just so happens that in the problem, x had an upper bound of 1. Would we have been out of luck if that hadn't been the case?
its a nasty integral u sure you are doing everything correctly? so the integral is 1/1+x^2 only?
 
  • #6
153
1
No, I was referring to Zondrinas trig sub suggestion!
 
  • #7
Zondrina
Homework Helper
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Aha, thanks. Obviously it just so happens that in the problem, x had an upper bound of 1. Would we have been out of luck if that hadn't been the case?
I don't see any discontinuity in the integrand at all, so I don't think it would matter.
 
  • #8
153
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Right right, we're using tan and not some other trig function. My bad again. Thanks a ton :)
 
  • #9
verty
Homework Helper
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Zondrina: Aha, thanks. Obviously it just so happens that in the problem, x had an upper bound of 1. Would we have been out of luck if that hadn't been the case?
That is a very good question, I don't know the answer to it. I'm going to investigate this.
 
  • #10
verty
Homework Helper
2,164
198
That is a very good question, I don't know the answer to it. I'm going to investigate this.
Sorry, the original question has no trouble because the range of tan is just R. Here is a more applicable problem:

##y = \int {dx \over (1 - x^2)^2}##

##x =? \; sin\theta##
##dx = cos\theta \; d\theta##

##y = \int sec^3\theta \; d\theta##
## = {1 \over 2} [sec\theta \; tan\theta + \int sec\theta \; d\theta] + C##
## = {1 \over 2} [sec\theta \; tan\theta + ln| sec\theta + tan\theta | ] + C##
## = {1\over 2}{x \over 1 - x^2} + {1\over 4} ln | {1+x\over 1-x}| + C##


On the other hand, one can make a translation:

##x = u-1##
##dx = du##

##y = \int {dx \over (1 - x^2)^2} = \int{ dx \over (1+x)^2 (1-x)^2} = \int {du \over u^2(2-u)^2}##

After some partial fraction magic:

##y = \int {1 \over 4(2-u)^2} + {1 \over 4u^2} + {1 \over 4u} + {1 \over 4(2-u)} du##
## = {1 \over 4} [ {1 \over 2-u} - {1\over u} + ln|u| - ln|2-u| ] + C##
## = {1 \over 4} [ {2(u-1) \over u(2-u)} + ln| {u \over 2-u} | ] + C##
## = {1 \over 2} {x \over 1 - x^2} + {1 \over 4} ln| {1+x \over 1-x} | + C##

I can't explain it but it's the same answer. Probably it has to do with the behaviour of sin(z) and cos(z) for complex z.
 

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