Integrate 1/(x^2-1): Solutions & Steps

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In summary, you can find the integral by breaking the fraction down into two parts and integrating by parts.
  • #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 I've 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 doesn't get anywhere either
 
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  • #2
Factor the denominator and split it in two, using partial fraction (decomposition).
 
  • #3
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
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
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 I am 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 doesn't 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
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
fizzzzzzzzzzzy said:
TD, i wasn't really taught formally about decomposition but from what I am 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 doesn't 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
 

1. What is the formula for integrating 1/(x^2-1)?

The formula for integrating 1/(x^2-1) is: ∫ 1/(x^2-1) dx = (1/2) ln |(x+1)/(x-1)| + C

2. How do I solve an integral with 1/(x^2-1) in the denominator?

To solve an integral with 1/(x^2-1) in the denominator, you can use the method of partial fractions. First, factor the denominator into (x+1)(x-1). Then, express 1/(x^2-1) as A/(x+1) + B/(x-1). Solve for the constants A and B, and then integrate each term separately.

3. Can I use substitution to solve an integral with 1/(x^2-1)?

Yes, you can use substitution to solve an integral with 1/(x^2-1). Substitute x = sinθ, dx = cosθ dθ, and use the trigonometric identity sin^2θ + cos^2θ = 1 to rewrite the integral in terms of θ. Then, use the formula for integrating sinθ/cosθ to solve for the integral.

4. Are there any special cases when integrating 1/(x^2-1)?

Yes, there are two special cases when integrating 1/(x^2-1). If the integral is of the form ∫ 1/(1-x^2) dx, you can use the substitution x = tanθ to solve it. If the integral is of the form ∫ 1/(x^2+1) dx, you can use the substitution x = tanhθ to solve it.

5. Can I use integration by parts to solve an integral with 1/(x^2-1)?

No, integration by parts cannot be used to solve an integral with 1/(x^2-1). This is because there is no algebraic manipulation that can be done to rewrite the integral in a simpler form. Instead, you must use one of the methods mentioned above, such as partial fractions or substitution.

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