the integral of 1/(1-x^2) (hyperbolic functions vs partial fractions?)

3 answers to the Integral of 1/(1-x^2), which one do we use?

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[ThePharphis]

response to Title:
I don't know but I'm such a noob I would probably attempt to use partial fractions

[aaronvanhorn]

dang, the kid sitting behind me in my calc class is going to be thanking me when I wear that derivatives shirt during a math test.

[SartajKhan-jgnz]

Fun thing, I was thinking about this just yesterday night. Today I see this! Cant explain the joy!

[XESolar]

I bought that shirt months ago and EVERYONE loves it!

[joao_pedro_c]

Please show us how to integrate from 0 to infinity 1/(1+x^2) dx with complex analysis?

[aram]

Could you consider another solution to be (tanh^-1x+coth^-1x)/2+C?

EDIT: actually, such a function would have no domain on the reals but its derivative is still 1/(1-x^2) o.o

[burgerhex]

For your question at the end (replacing 1 - x^2 with a^2 - x^2), I got this result:

∫1/(a^2-x^2) =

1/(2a) * ln((a+x)/(a-x))

[aadiduggal]

Hey, can you do a proof for the taylor/power series? My calculus textbook just states it, and the proofs online are pretty confusing. How do we know its true?

Also, do you teach at Berkeley?

[asusmctablet]

So does that mean that tanh^-1(x) equals 1/2*ln((1+x)/(1-x)) for -1<x<1?

[timlu]

I saw an integral problem on a website and have been trying it a year ago but still can't solve it:

I(n) = 0->π ∫sin^2(nx)/sin^2(x) dx for all integers n.

I wish Mr. Cao or viewers can solve this in the term of n and reply to this comment. Thanks!

[crismikelarciga]

You can also use trig substitution and will ended up integrating sec(theta) d(theta) and there can be other answers

[janderson]

That’s super weird. I was struggling with this problem just last week

[bjorn]

you can also use trig sub, let x = sinθ. And i got the same answer from soln#3

[meowing]

Love the background piano doreamon song in the beginning

[JeffreyLByrd]

I would’ve done partial fractions in the first place, because I would’ve never thought of the hyperbolic functions.

[sultanelshirazy]

I always use partial fraction because I haven't learned about inverse nor hyperbolic functions

[dannyundos]

Funny thing is, none is a constant shift of another, even in complex domain. Is this because of the poles?

[physicsphysics]

CORRECTIONS:
7:27 If the domain excludes just a point, you usually don't have to include the if.

[Piemaster]

Ok, so this is really picky, but the domain considerations bring up a second technical consideration around the constant of integration that you are ignoring here. Because the domains of these functions (except tanh^{-1}(x)) are disconnected, a full picture of the antiderivative of 1/(1-x^2) would have potentially different constants for each connected component in the domain.

As an example, if I asked for the function f(x) where f'(x) = 1/(1-x^2) and f(0) = 4, your answer might be tanh^{-1}(x) + 4 or 1/2 ln|(1+x)/(1-x)| + 4 which are correct answers. However, these are only two of infinitely many possible correct answers to this question (making the question somewhat ill-formed). If I want there to be a single answer, I should specify function values over each connected component in the domain of 1/(1-x^2), meaning I would need 3 values specified. As an example, I could ask for f(x) where f'(x) = 1/(1-x^2) and, f(-2)=0, f(0)=4, and f(5)=-1. Then, the correct answer for f(x) would be different depending on the domain. If x<-1, you would get f(x) = 1/2 ln|(1+x)/(1-x)| + 1/2 ln(3). If -1 < x < 1, you would get f(x) = 1/2 ln|(1+x)/(1-x)| - 4. If x>1, you would get f(x) = 1/2 ln|(1+x)/(1-x)| -1 -1/2 ln(3/2). These three pieces specify a single piecewise-defined function which completely characterizes the specific antiderivative in question.

The reason this is so picky is that when doing definite integrals, the constants do not matter. Since most people doing integrals are trying to evaluate definite integrals, many professors will ignore this technicality for the sake of avoiding unnecessary complexity when presenting it to students. Where this becomes vital is in an advanced functional analysis or differential geometry course when examining function spaces over the real line, so basically math grad school. The fact that you can have different constants over each connected component of the domain exhibits the fact that the 0-dimensional de Rham cohomology is a real vector space with dimension equal to the number of connected components in your domain.

[alekisighl]

I just noticed one thing if you graph the three functions you can see that:
If |x| <1 then 1/2 ln(|1+x/1-x|) = tanh^-1(x)
And if |x| >1 then 1/2 ln(|1+x/1-x|) = coth^-1(x)
Why is this?