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

• fizzzzzzzzzzzy
In summary, you can find the integral by breaking the fraction down into two parts and integrating by parts.

#### fizzzzzzzzzzzy

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

Factor the denominator and split it in two, using partial fraction (decomposition).

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.

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?

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.

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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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?

It looks to me that the best method is to use partial fractions as mentioned.

$$\frac{1}{x^2-1}$$

Then make x^2-1 into two terms

$$\frac{1}{(x+1)(x-1)}$$

$$\frac{1}{x+1}.\frac{1}{x-1}$$

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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Alternatively, use the substitution:
$$x=Tanh(u)\to{dx}=\frac{du}{Cosh^{2}(u)}=du(1-Tanh^{2}(u))=(1-x^{2})du$$

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:

$$\frac{1}{{x^2 - 1}} = \frac{1}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = \frac{1}{{2\left( {x - 1} \right)}} - \frac{1}{{2\left( {x + 1} \right)}}$$

This last expression is easy to integrate.

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.