Integration problem

  • #1
im trying to figure out the integral of 1/(x^2-1).
i basically have no clue how to do this, since every thing ive tried has resulted in a dead end
i tried intergration by parts by factoring the denominator but that just results in a more complex integral.
i also tried tried multiplying both sides of the fraction by 2x and then substituting x^2-1 with u but that doesnt get anywhere either
 

Answers and Replies

  • #2
TD
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Factor the denominator and split it in two, using partial fraction (decomposition).
 
  • #3
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i think this does work by integration by parts

show us what you did and maybe someone with more knowledge can answer you or point out where you're going wrong.
 
  • #4
alright, what i did when i tried to integrate it by parts is split the denominator
(1/x-1)*(1/x+1) then i took 1/(x-1) as u and 1/(x+1) as dv. then the new intergral i have to solve is vdu, or the integral of (x-1)^(-2)*ln(x+1) well in this case, ill have to integrate by parts again, but if i choose ln(x+1) as u, then i just get the integral i had before and if i choose (x-1)^(-2) as u, the integral just becomes more complex. did i do sometime wrong?
 
  • #5
TD
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I really don't see why you'd want to use integration by parts here.
Try my suggestion, it's obviously the standard approach for this one.
 
  • #6
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You can do it by parts and it should be done to make it a little easier and aesthetically pleasing =). You just need to setup a triangle. Draw a triangle first, label each side accordingly.

After that, you can get your trig functions out of it. Then you can take your original x that you solved for and put it in.

You need to look for a few things to replace: the dx (which you will solve for x, differentiate that and you get your dx to replace that, then you can plug that back in the end [what x equals]). The denominator part and/or the numerator.
 
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  • #7
TD, i wasn't really taught formally about decomposition but from what im seeing on wikipedia, its basically factoring the denominator, then using substitution right? but if i do that, i just end up with 1/u*(u-2) or 1/u*(u+2), which doesnt help me much. could you maybe show what you mean by decomposition if this wasnt what you meant?
 
  • #8
It looks to me that the best method is to use partial fractions as mentioned.

[tex] \frac{1}{x^2-1}[/tex]

Then make x^2-1 into two terms

[tex]\frac{1}{(x+1)(x-1)}[/tex]

[tex]\frac{1}{x+1}.\frac{1}{x-1}[/tex]

Then proceed from there, you don't really need substitution.

I checked the answer on a maths program and it came up with -atanh(x) hehe, I presume it's equivalent. your answer may well look differrent from this and involve logs.:/
 
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  • #9
arildno
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Alternatively, use the substitution:
[tex]x=Tanh(u)\to{dx}=\frac{du}{Cosh^{2}(u)}=du(1-Tanh^{2}(u))=(1-x^{2})du[/tex]
 
  • #10
TD
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TD, i wasn't really taught formally about decomposition but from what im seeing on wikipedia, its basically factoring the denominator, then using substitution right? but if i do that, i just end up with 1/u*(u-2) or 1/u*(u+2), which doesnt help me much. could you maybe show what you mean by decomposition if this wasnt what you meant?
You write the fraction as a sum of fractions with a lineair denominator:

[tex]
\frac{1}{{x^2 - 1}} = \frac{1}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = \frac{1}{{2\left( {x - 1} \right)}} - \frac{1}{{2\left( {x + 1} \right)}}
[/tex]

This last expression is easy to integrate.
 
  • #11
oh wow, i feel pretty stupid that i didnt see that before. thanks for the help
 

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