# Intergration problem

1. Feb 8, 2009

### miller8605

1. The problem statement, all variables and given/known data
i'm taking a defitinite integral from sqrt2 to 2 of the function 1/x^3*sqrt(x^2-1)dx.

2. Relevant equations

3. The attempt at a solution
I seperated it into 1/x^3 and 1/sqrt(x^2-1). I have the second part using trig sub. as being sec theta dtheta, before integrating it. I believe i did this part correctly.

What I can't remember is that I make 1/x^3 to x^-3 and then integrate it that way with the final being -1/2(1/x^2)??

2. Feb 8, 2009

### Dick

I know the forums are slow, but you can stop posting this now. The substitution you want is u^2=x^2+1. Work it out. You can turn it into a rational function in u. Then use partial fractions.

3. Feb 8, 2009

### miller8605

for some reason this forum isn't working, and i didn't mean to repost several times... I hit refresh a couple times and that's what happened. I apologize!

4. Feb 8, 2009

### Dick

S'ok. Can you remove the other ones or mark as duplicates?

5. Feb 8, 2009

### miller8605

Ok, was i correct in splitting it into two integrals?

I'm having alot of difficulty on this one. I got an answer, but it's really long and seems wrong.

I found I'm supposed to get but need to be able to work through this problem.

How i got mine was for the second part used x = sec y and dx= sec y tan y dy.
after working the substitution i came up with the integral of sec y dy from sqrt(2) to 2. the integration of sec is ln abs(sec y + tan y). I then substituted the inverse sec (x/a) in for y based on my original substitution.

that is then multiplied by the integral of 1/x^3 which is -1/(2x^2). Was this the correct way in doing it?

6. Feb 8, 2009

### Dick

You can't 'split it into two integrals'. There's no rule that says integral(f*g)=integral(f)*integral(g). That's just plain wrong. You can probably handle it with a trig substitution as well, but I would suggest using my original suggestion of u^2=x^2+1.

7. Feb 8, 2009

### miller8605

ok. do you mean to say substitute u^2=x^2 -1 to get rid of the sqrt then?

i just don't understand how that helps me. I end up with variables x and u in the denominator then.

8. Feb 8, 2009

### Dick

Yes, I meant to say u^2=x^2-1. Sorry. It still works. The stray x will cancel out. x^2=u^2+1.

Last edited: Feb 9, 2009
9. Feb 9, 2009

### miller8605

what would x^3 be in terms of u? I'm gonna have to head to a tutor and have someone walk me through this one.

10. Feb 9, 2009

### AssyriaQ

Here is a hint: don't express x3 in terms of u. At first, express only the square root term and dx in terms of u and du. This will give you an easier term than x^3 to express in terms of u.