Solve Integral: \int \frac{x^2-1}{(x^2+1) \sqrt{x^4+1} }\ \mbox{d}x

[dirk_mec1]

$$\int \frac{x^2-1}{(x^2+1) \sqrt{x^4+1} }\ \mbox{d}x$$


I know I have to use some substitution but I can't figure out which one can someone help me?

 

[Defennder]

Try it out on the online integrator, there's no elementary solution for the indefinite integral.

 

[Gib Z]

Sometimes the Integrator gets it wrong. If the OP knows for a fact it is do-able, we should wait a little longer before dismissing it. Perhaps he missed limits on the integral. At first sight I might try somethings with u=x^2+1.

 

[dirk_mec1]

Sometimes the Integrator gets it wrong. If the OP knows for a fact it is do-able, we should wait a little longer before dismissing it. Perhaps he missed limits on the integral. At first sight I might try somethings with u=x^2+1.

Strange... indeed the online integrator gives an elliptic integral. By the way I found out that they say use

$$u = \sqrt{x^2 +x^{-2}}$$

But I can't simplify the integral.

 

[fantispug]

Strange... indeed the online integrator gives an elliptic integral. By the way I found out that they say use

$$u = \sqrt{x^2 +x^{-2}}$$

But I can't simplify the integral.

I don't think I would have ever thought of that substitution personally, but it works like a charm. Where do you get stuck?

 

[dynamicsolo]

I don't think I would have ever thought of that substitution personally, but it works like a charm. Where do you get stuck?

This is an interesting substitution, and du alone gets you

$$du = \frac{x^4-1}{\sqrt{x^4+1} }\ \mbox{d}x$$.

That $$x^4 - 1$$ in the numerator would be perfect, if we could just get
an $$(x^2 + 1)^{2}$$ into the denominator. I'm not seeing how that's going to happen...

At first sight I might try somethings with u=x^2+1.

Alas, there are a number of substitutions that get one tantalizingly close. I also tried rationalizing the denominator of the integrand first as one tack and tried using the substitution $$tan \theta = x^2$$ as another. If the differentials cooperate, then one of the factors doesn't, and vice versa...

 

[Gib Z]

Ahh After I get that (x^2+1)^2 term in the denominator in terms of you, I end up with something like $$\int \frac{1}{(u^2 + \sqrt{u^4-4} + 2)^2} du$$.

It doesn't look like that will turn out nice.

 

[dirk_mec1]

I don't think I would have ever thought of that substitution personally, but it works like a charm. Where do you get stuck?

Really? Works like a charm? Please show me I get the same problem as dynamicsolo.

@Gib Z how did you get that form of the integral?

 

[dirk_mec1]

With huge help form an expert:

The integrand is $$\frac{x^2 -1 }{(x^2 + 1) \sqrt{x^4 + 1}}.$$
We then "extract" x^2 from the second term in the denominator to get$$\frac{x^2 - 1}{x(x^2 + 1) \sqrt{x^2 + 1/x^2}}$$, which can then be written as$$\frac{1 - 1/x^2}{(x + 1/x) \sqrt{(x + 1/x)^2 - 2}}.$$
Now use the substitution t = x + 1/x and not that dt = (1 - 1/x^2) dx to find:$$\int \frac{dt}{t (t^2-2)}$$

Use $$t= \sqrt{2} \sec( \theta)$$ to find:$$I = \frac{1}{\sqrt{2}} \cdot \int\ 1\ \mbox{d} \theta = \frac{1}{\sqrt{2}} \theta +C$$Now can someone explain to me why the online integrator gives an incorrect answer!

 

[m_s_a]

http://www.up07.com/up7/uploads/4eeb460777.jpg

 

[dirk_mec1]

So the online integrator splits the integral and encounters an integral that cannot be expressed in elementary functions?

 

[m_s_a]

I do not know

 

[dynamicsolo]

It strikes me that integration software works in a similar way to translation software: it finds a particular way to parse a statement and then proceeds to work with the decomposition it created. If the choice made by the analysis strategy is a poor one, the translation will be rather weak. (Analogously, the integrand may be broken into forms of which some give unsatisfactory results.) The options will be limited by the situations that the programmers considered in preparing the code.

I've seen a page of integration problems and solutions given to students by one instructor where they used software to provide the answers. A few of the solutions were a little... odd. We've run into similar integration difficulties on other occasions on this Forum when someone tried an exotic integral with software...

With huge help form an expert: ...

Wow, my hat is off to the person who spotted that approach!

This line should be

$$\int \frac{dt}{t \sqrt{t^2-2}}$$ ,

though, shouldn't it?

 

[fantispug]

Really? Works like a charm? Please show me I get the same problem as dynamicsolo.

@Gib Z how did you get that form of the integral?

You do a similar thing to factoring out the x^2 to what dynamicsolo did in their final solution:

As
$$\ \mbox{d}u = \frac{x^4-1}{ \sqrt{x^4+1} }\ \mbox{d}x = \frac{x^2-1}{\sqrt{x^4+1}}\frac{x^2+1}{x^2}\ \mbox{d}x$$
Substituting this into the integral:
$$\int{\frac{x^2-1}{\left(x^2+1\right)\sqrt{x^4+1}}\ \mbox{d}x}=\int{\frac{x^2}{\left(x^2+1\right)^2}\ \mbox{d}u}$$
Expand and simplify the bottom line to find a pleasant surprise, and an equivalent answer.

 

[Gib Z]

This line should be

$$\int \frac{dt}{t \sqrt{t^2-2}}$$ ,

though, shouldn't it?

Yes it should, I got that one as well before but it didn't turn out so good.

 

[dirk_mec1]

Yes it should, I got that one as well before but it didn't turn out so good.

Yes you're correct my bad.

@fantisplug what happened in the second Latex formula?

 

[dynamicsolo]

$$\ \mbox{d}u = \frac{x^4-1}{ \sqrt{x^4+1} }\ \mbox{d}x = \frac{x^2-1}{\sqrt{x^4+1}}\frac{x^2+1}{x^2}\ \mbox{d}x$$

The numerator is, of course, a difference of two squares, and factors as you've shown it. Where does the x^2 in the denominator come from (without affecting the radical at all)...?

 

[ace123]

The integrand is $$\frac{x^2 -1 }{(x^2 + 1) \sqrt{x^4 + 1}}.$$



We then "extract" x^2 from the second term in the denominator to get


$$\frac{x^2 - 1}{x(x^2 + 1) \sqrt{x^2 + 1/x^2}}$$, which can then be written as


$$\frac{1 - 1/x^2}{(x + 1/x) \sqrt{(x + 1/x)^2 - 2}}.$$



Does my algebra suck or is that not possible? How would extract the x^2 from the second term. Wouldn't that change it?

Basically what dynamicsolo asked.

 

[Gib Z]

We divided both numerator and denominator by x^2. For the denominator that meant dividing each factor by x. Which, when dividing the sqrt, is dividing by sqrt(x^2).

 

[dynamicsolo]

I think this has been a case of everyone not showing important steps, making algebra errors or typos in key places, and not spotting other people's mistakes (myself and my own included...).

As a reminder, here's the integral:

$$\int \frac{x^2-1}{(x^2+1) \sqrt{x^4+1} }\ \mbox{d}x$$

In discussing the substitution

$$u = \sqrt{x^2 +x^{-2}}$$ ,

I flubbed the distribution of factors in the denominator in reducing the radical, so my result in post #6 is wrong; I now agree with fantispug in the result for du. However, fantispug in turn made a typo in the first equation in post #14, which should read:

$$\ \mbox{d}u = \frac{x^4-1}{<b>x^2</b> \sqrt{x^4+1} }\ \mbox{d}x = \frac{x^2-1}{\sqrt{x^4+1}}\frac{x^2+1}{x^2}\ \mbox{d}x$$ ;

the result is correct, but the typo made in the first half of the equation made it look like there was a factorization error.

I'm still unclear for this substitution though, fantispug, how the remaining segment of this integrand turns out nicely in terms of u:

$$\int{\frac{x^2}{\left(x^2+1\right)^2}\ \mbox{d}u}$$

It still seems like there is an excess term left over when transforming from a function of x to a function of u... May I ask that you show this?

It also seems that we never actually said what the integral is. Using the substitution in post #9 that dirk_mec1's expert provided:

t = x + 1/x ,

I agree with the intermediate result

$$I = \frac{1}{\sqrt{2}} \theta +C$$ .

Since theta is related to an arcsecant, I figure it is better to square the trig substitution relation first:

$$sec^{2} \theta = \frac{t^2}{2}$$

and use the Pythagorean Identity to get

$$tan^{2} \theta = \frac{t^2 - 2}{2}$$ .

We already used $${t^2 - 2} = \frac{x^4 + 1}{x^4}$$ , so we have

$$tan\theta = \sqrt{\frac{x^4 + 1}{2x^4} }$$,

giving us

$$I = \frac{1}{\sqrt{2} } arctan[ \sqrt{\frac{x^4 + 1}{2x^4} } ] + C$$ .

You could also flip over the secant-squared expression to make it cosine-squared, use the Pythagorean identity to get

$$sin^{2} \theta = \frac{t^2 - 2}{t^2}$$ ,

leading to

$$sin \theta = \frac{\sqrt{x^4 + 1} }{x^2 + 1}$$

and thus.

I = \frac{1}{\sqrt{2} } arcsin[ {\frac{\sqrt{x^4 + 1} }{x^2 + 1} } ] + C [/tex] .

(And I think I caught all of my mistakes this time...)

In any case, I think it's pretty clear that the integration software does not have an appropriate strategy for dealing with this little problem...

 

[Littlepig]

In any case, I think it's pretty clear that the integration software does not have an appropriate strategy for dealing with this little problem...

Definitely don't use wolfram to integrate, that's a BIG mistake. I use mathematica 6 and there is a lot to improve in integration...It is normal to give something impossible or something non-sense(like elliptical or hyperbolic stuff) with moderate integrals, specially with functions like this one.;)

However, that is definitely a nice integral to pratice..:P

 

[dynamicsolo]

Part of this still isn't right (and I ran out of my 30 minutes of "edit time"...).

...

t = x + 1/x ,

...

We already used $${t^2 - 2} = \frac{x^4 + 1}{x^4}$$

That should be $${t^2 - 2} = \frac{x^4 + 1}{x^2}$$,
so the integral in this form should be

$$I = \frac{1}{\sqrt{2} } arctan[ \sqrt{\frac{x^4 + 1}{2x^2} } ] + C$$ .

As for the other form using arcsin,

$$sin^{2} \theta = \frac{t^2 - 2}{t^2}$$ ,

leading to

$$sin \theta = \frac{\sqrt{x^4 + 1} }{x^2 + 1}$$

I think I managed to get that right. So this version of the integral would be

$$I = \frac{1}{\sqrt{2} } arcsin[ {\frac{\sqrt{x^4 + 1} }{x^2 + 1} } ] + C$$ .

Anyone spot any further mistakes? Maybe we can put this one to rest at last...

 

[Gib Z]

Differentiating returns the integrand, after about half a page of working lol. Nice Solution.

 

[Defennder]

Yes I agree it's surprising. Someone should inform the Mathematica development team on this.

 

[dynamicsolo]

Yes I agree it's surprising. Someone should inform the Mathematica development team on this.

Apropos of a comment I made earlier in this thread, I think the problem of creating a general integrator program is comparable in difficulty to that of a language translation program. There is something about the ingenuity of an experienced human mind in finding solutions to integrals such as this one that seems difficult to capture in an algorithmic procedure.

I'd read recently about an encounter of the young Jeremy Bernstein with John von Neumann (in what had to be the early days of electronic computers and cybernetics, given that "Johnny" was still alive), when there was an effort to develop programs that could enable computers to work out mathematical proofs (and, in fact, there was some success in getting computers to do geometric "reasoning"). Bernstein asked, "Do you think computers will ever replace mathematicians?", to which von Neumann, chuckling, replied, "Sonny, I wouldn't worry about that!"

 

[dynamicsolo]

I couldn't quite leave this integral alone yet, because when I first graphed the function [ http://usera.imagecave.com/dynamicsolo/PFpix/integral0608a.jpg ], I was curious about what the area under the curve would turn out to be. Of course, because it is a function with even symmetry, we expect

$$\int_{-\infty}^{\infty} \frac{x^2-1}{(x^2+1) \sqrt{x^4+1} }\ \mbox{d}x = 0$$ .

But once we had the general anti-derivative, it looked like it might also be the case that

$$\int_{0}^{\infty} \frac{x^2-1}{(x^2+1) \sqrt{x^4+1} }\ \mbox{d}x = 0$$ ,

since the values of either form of the antiderivative are the same at zero and (in the limit at) infinity.

I originally looked at this because I thought there was some strange "kink" about the substitutions that were used that was going to require a sign change in the antiderivative in certain intervals. This turned out not to be the case... Here's a graph [ http://usera.imagecave.com/dynamicsolo/PFpix/integral0608b.jpg ] illustrating the substitutions that were used:

t = x + 1/x and $$t = \sqrt{2} sec\theta$$ .

The curious thing about the transformation is that the entire right half of the x-axis gets mapped into $$[ \frac{\pi}{4} , \frac{\pi}{2} )$$ for theta, and the mapping folds back on itself. This leads to an interesting result for the improper integral.

When you graph the arguments for the inverse trig functions that are the antiderivatives found, you get a similar sort of folding:

for $$arctan[ \sqrt{\frac{x^4 + 1}{2x^2} } ]$$ -- http://usera.imagecave.com/dynamicsolo/PFpix/integral0608c.jpg

for $$arcsin[ {\frac{\sqrt{x^4 + 1} }{x^2 + 1} } ]$$ -- http://usera.imagecave.com/dynamicsolo/PFpix/integral0608d.jpg

Using either form of the antiderivative, we find that

$$\int_{0}^{1} \frac{x^2-1}{(x^2+1) \sqrt{x^4+1} }\ \mbox{d}x = <br /> <br /> \frac{1}{\sqrt{2}} [ \frac{\pi}{4} - \frac{\pi}{2} ] = -\frac{\pi\sqrt{2}}{8}$$

and

$$\int_{1}^{\infty} \frac{x^2-1}{(x^2+1) \sqrt{x^4+1} }\ \mbox{d}x = <br /> <br /> \frac{1}{\sqrt{2}} [ \frac{\pi}{2} - \frac{\pi}{4} ] = \frac{\pi\sqrt{2}}{8}$$ .

Thus the region of negative signed area between x = 0 and x = 1 exactly cancels the positive signed area for x > 1 . The two slim infinite wings of the integrand have the same area as the large bulge below the x-axis.

As a check for the credibility of the value of the area obtained here, I have marked in the close-up graph of the integrand (lower part of first graph) a red line from the minimum at (0, -1) to the zero at (1, 0). The area of the triangle formed by that segment and the coordinate axis is, of course, 1/2 , while the absolute value of the area of the "bulge" from x = 0 to x = 1 is $$\frac{\pi\sqrt{2}}{8} \approx 0.5554$$ , which looks believable on the graph.