Need help with an anti-derivative

[PowerMacG5]

I am trying to find the anti-derivative of
int(1/sqrt(x^4+1))dx from 1 to infinity

I already tried substituting u^2=x^4 + 1 but I eventually hit a dead end. I also tried calc101.com but it cannot evaluate the integral.

Also, can someone help me with the limit:
lim((2x-sin(x))/(3x+sin(x)) as x approaches infinity.
I already tried L'Hôpital's rule, but that still gives me the limit:
lim((2-cos(x))/(3+cos(x)) as x approaches infinity. As x approaches infinity cos(x) varies between -1 and 1, and therefore the limit is undefined.
BTW, my Ti-89 evaluates the original limit as 2/3, but the new one after L'Hôpital as undefined.

 

[beer]

nm

 

[ClueLis]

Substitute u=x^2. This will leave you with 1/(2sqrt(u)sqrt(u^2+1)). Break it up using partial fraction decomposition to obtain 1/2(A/sqrt(u)+B/sqrt(u^2+1)), which will be much easier to integrate.

Edit: Hmmm... I can't get the fractions to separate the way I want them to.

 

[z0mb13]

why didnt u just use integral??

isnt it the same thing??

 

[PowerMacG5]

Originally posted by: ClueLis
Substitute u=x^2. This will leave you with 1/(2sqrt(u)sqrt(u^2+1)). Break it up using partial fraction decomposition to obtain 1/2(A/sqrt(u)+B/sqrt(u^2+1)), which will be much easier to integrate.

Okay, when I try that, I get
1=A(sqrt(u^2+1))+B(sqrt(u))

I cannot figure out how to solve for A and B.

 

[ClueLis]

Originally posted by: Marauder911

Originally posted by: ClueLis
Substitute u=x^2. This will leave you with 1/(2sqrt(u)sqrt(u^2+1)). Break it up using partial fraction decomposition to obtain 1/2(A/sqrt(u)+B/sqrt(u^2+1)), which will be much easier to integrate.

Okay, when I try that, I get
1=A(sqrt(u^2+1))+B(sqrt(u))

I cannot figure out how to solve for A and B.

That's what I just noticed, too.

 

[PowerMacG5]

Originally posted by: ClueLis

Originally posted by: Marauder911

Originally posted by: ClueLis
Substitute u=x^2. This will leave you with 1/(2sqrt(u)sqrt(u^2+1)). Break it up using partial fraction decomposition to obtain 1/2(A/sqrt(u)+B/sqrt(u^2+1)), which will be much easier to integrate.

Okay, when I try that, I get
1=A(sqrt(u^2+1))+B(sqrt(u))

I cannot figure out how to solve for A and B.

That's what I just noticed, too.

Ah, ok. Anyone else have any suggestions?

 

[Lynx516]

I cannot solve it and nor can my Ti-89. That is one hard problem

 

[PowerMacG5]

Originally posted by: Lynx516
I cannot solve it and nor can my Ti-89. That is one hard problem

I know the Ti-89 cannot, neither can calc101. I have a feeling that it is a simple problem after you make some sort of observation. If the numerator was just x^3, then it would be simple. If it helps, I am supposed to evaluate the integral from 1 to ininity (I just added that to my original post), maybe that would make it easier. When I plug that into maple I get (1/2)EllipticK((1/2)sqrt(2))

According to the documentation, EllipticK is a complete elliptical integral of the first kind, and I have no idea what that means.

 

[CrackaLackaZe]

For the first antider problem, did you try using trig sub? there should be a trig sub that corresponds to
int[sqrt(a^2+u^2)], a being x^2 and u being 1. I think the sub is u = a tan theta. I'm not sure if thats the right sub cause its been awhile, but I'm pretty sure thats the easiest way to evaluate that integral.

Sorry its easier with u = x^2 and a = 1.

 

[VTboy]

Look at all of the people helping you. Not fair. I post simple algebra and DE problems and hardly anyone can help me.

 

[makessense]

Originally posted by: CrackaLackaZe
For the first antider problem, did you try using trig sub? there should be a trig sub that corresponds to
int[sqrt(a^2+u^2)], a being x^2 and u being 1. I think the sub is u = a tan theta. I'm not sure if thats the right sub cause its been awhile, but I'm pretty sure thats the easiest way to evaluate that integral.

Sorry its easier with u = x^2 and a = 1.

what i was thinking.

 

[PowerMacG5]

Originally posted by: CrackaLackaZe
For the first antider problem, did you try using trig sub? there should be a trig sub that corresponds to
int[sqrt(a^2+u^2)], a being x^2 and u being 1. I think the sub is u = a tan theta. I'm not sure if thats the right sub cause its been awhile, but I'm pretty sure thats the easiest way to evaluate that integral.

Sorry its easier with u = x^2 and a = 1.

Have you tried what you suggested? Because I have, and this is what i get:

int(1/sqrt(1+x^4))dx
a=1
u = x^2
du = 2xdx

(1/2)int(1/(x*sqrt(a+u^2)))du = (1/2)int(1/(sqrt(u)*sqrt(a+u^2)))du which brings us back to the partial fractions. which brings us back to the wall.
Even if I continue after this, I get this:
u = a*tan(theta)
du = a*sec(theta)^2 d(theta)

(1/2)int((a*sec(theta)^2)/(sqrt(a*tan(theta))*sqrt(a^2*sec(theta)^2)))d(theta) =
(1/2)int(sec(theta)/sqrt(tan(theta)))d(theta) (I substituted 1 back in for a to simplify it)
Now what do I do? It is still an integral that my calculator cannot solve, nor can I.

 

[CrackaLackaZe]

int[1/sqrt(x^4+1) dx]: sqrt(u^2+a^2) [a = 1, u = x^2]

[u= a tan theta, thus x^2= (1) tan theta

x^2=(1)tan theta, x = sqrt[tan theta], dx = (1/2)(1/sqrt(theta))(sec^2 theta) d theta]

= 1/2 int[[sec^2 theta]/[(sqrt(tan theta))(sec^2 theta)] d theta (tan^2 theta + 1 = sec^2 theta)

= 1/2 int (1/sqrt(tan theta) d theta

 

[PowerMacG5]

Originally posted by: CrackaLackaZe
int[1/sqrt(x^4+1) dx]: sqrt(u^2+a^2) [a = 1, u = x^2]

u= a tan theta, thus x^2= (1) tan theta

x^2=(1)tan theta, x = sqrt[tan theta], dx = (1/2)(1/sqrt(theta))(sec^2 theta) d theta

= 1/2 int[[sec^2 theta]/[(sqrt(tan theta))(sec^2 theta)] d theta

= 1/2 int (1/sqrt(tan theta) d theta

This isn't correct because you cannot just say u=x^2 and then say du=dx because that's not correct. du=2x dx which is what complicates everything. Also, its sqrt(sec(theta)^2) on the bottom, so you still are left with a sec(theta) on the top.

 

[PowerMacG5]

Bump

 

[PowerMacG5]

Bump for the night people.

 

[Aves]

Do we know for sure there is an antiderivative for that function? Why can't we just compare it to a function like 1/x, say it's divergent at leave it at that? 😛

 

[jessicak]

which math book do you use?

 

[DrPizza]

lim((2x-sin(x))/(3x+sin(x)) as x approaches infinity.

For what it's worth, just from a quick inspection, the limit is 2/3. I'm not sure what work you need to show to support the answer. (simply think of a huge number. You have 2 times the huge number over 3 times the huge number. The value of sinx will vary from -1 to 1 and is of no significance.

 

[Legendary]

Originally posted by: DrPizza

lim((2x-sin(x))/(3x+sin(x)) as x approaches infinity.

For what it's worth, just from a quick inspection, the limit is 2/3. I'm not sure what work you need to show to support the answer. (simply think of a huge number. You have 2 times the huge number over 3 times the huge number. The value of sinx will vary from -1 to 1 and is of no significance.

Yep, at extremely large numbers (like infinity) variations of 1 and -1 make no difference.
Just confirming Dr. Pizza. Too lazy to do the integral.

 

[PowerMacG5]

Originally posted by: aves2k
Do we know for sure there is an antiderivative for that function? Why can't we just compare it to a function like 1/x, say it's divergent at leave it at that? 😛

LOL, that's what the teacher ended up doing today when he went over it, although the function actually converges (it isn't divergent), but he said he even could not find out what it converges to. I knew it converged because the Ti-89 gives a decimal approximation, and maple gives it in terms of the EllipticK function. He compared it to the function 1/sqrt(x^4). No one in my class was able to get the problem because he never went over comparisons like this (it happens to be in the next chapter), we all tried everything to evaluate the integral. I have about 4 pages of work just trying to get it based on all the suggestions here.

Originally posted by: jessicak
which math book do you use?

Calculus: Brief Edition, Seventh Edition by Anton, Bivens, and Davis.

Originally posted by: DrPizza

lim((2x-sin(x))/(3x+sin(x)) as x approaches infinity.

For what it's worth, just from a quick inspection, the limit is 2/3. I'm not sure what work you need to show to support the answer. (simply think of a huge number. You have 2 times the huge number over 3 times the huge number. The value of sinx will vary from -1 to 1 and is of no significance.

Thanks for the help. The way my teacher did it was he just divided both the numerator and denominator by x, so you get (2-sin(x)/x)/(3+sin(x)/x) and the limit of sin(x)/x as x approaches infinity is zero, so the limit approaches 2/3. Thanks everyone for the help.

 

[Aves]

Originally posted by: Marauder911

Originally posted by: aves2k
Do we know for sure there is an antiderivative for that function? Why can't we just compare it to a function like 1/x, say it's divergent at leave it at that? 😛

LOL, that's what the teacher ended up doing today when he went over it, although the function actually converges (it isn't divergent), but he said he even could not find out what it converges to.

You're right. I was in a hurry and didn't give it much thought but 1/x is greater than 1/sqrt(x^4+1) so the fact that 1/x diverges doesn't mean squat.

What was the funtion that the teacher used for the comparison?

 

[PowerMacG5]

Originally posted by: aves2k

Originally posted by: Marauder911

Originally posted by: aves2k
Do we know for sure there is an antiderivative for that function? Why can't we just compare it to a function like 1/x, say it's divergent at leave it at that? 😛

LOL, that's what the teacher ended up doing today when he went over it, although the function actually converges (it isn't divergent), but he said he even could not find out what it converges to.

You're right. I was in a hurry and didn't give it much thought but 1/x is greater than 1/sqrt(x^4+1) so the fact that 1/x diverges doesn't mean squat.

What was the funtion that the teacher used for the comparison?

Read my entire post, I say 1/sqrt(x^4).

 

[LordMorpheus]

Originally posted by: Marauder911

Originally posted by: ClueLis

Originally posted by: Marauder911

Originally posted by: ClueLis
Substitute u=x^2. This will leave you with 1/(2sqrt(u)sqrt(u^2+1)). Break it up using partial fraction decomposition to obtain 1/2(A/sqrt(u)+B/sqrt(u^2+1)), which will be much easier to integrate.

Okay, when I try that, I get
1=A(sqrt(u^2+1))+B(sqrt(u))

I cannot figure out how to solve for A and B.

That's what I just noticed, too.

Ah, ok. Anyone else have any suggestions?

Well, let U=0, that meanst that A must equal 1.

hmmmmm. That means that B varies with U.
edit: does this integral even converge?

1/x doesn't. 1/x^2 DOES converge, though, and x^2 = sqrt(X^4)


you know, if the original X^4 had been X^5, this would have been many, many, many times easier.

What chapter are you covering? This might be easier if i knew to seperate the fractions or use parts or something like that.