# Integrating Definite Integral: sqrt2 to 2 1/x^3*sqrt(x^2-1)dx

• miller8605
In summary, the conversation is about finding the definite integral of the function 1/x^3*sqrt(x^2-1)dx from sqrt2 to 2. The original poster tried separating the integral into two parts, but was advised to use the substitution u^2=x^2+1 instead. They are struggling with the substitution and how to express x^3 in terms of u. They are advised to express the square root term and dx in terms of u and du first.
miller8605

## Homework Statement

i'm taking a defitinite integral from sqrt2 to 2 of the function 1/x^3*sqrt(x^2-1)dx.

## The Attempt at a Solution

I separated 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)??

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.

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!

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

Dick said:
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.

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

I'm having a lot 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?

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.

Dick said:
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.

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.

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:
Dick said:
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.

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

miller8605 said:
what would x^3 be in terms of u?
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.

## 1. What is the purpose of integrating a definite integral?

The purpose of integrating a definite integral is to find the exact value of the area under a curve between two given points. It is a fundamental tool in calculus and is used to solve a variety of practical problems in physics, engineering, and other fields.

## 2. What does "sqrt2 to 2" represent in the given integral?

"sqrt2 to 2" represents the limits of integration, or the two points between which the area under the curve is being calculated. In this case, it means that the integral is being evaluated from the point x=sqrt2 to x=2.

## 3. What does the function 1/x^3*sqrt(x^2-1) represent?

The function 1/x^3*sqrt(x^2-1) represents the curve whose area is being calculated. It is a rational function that includes a square root term and is therefore a type of transcendental function.

## 4. How is the definite integral of a rational function with a square root term evaluated?

The definite integral of a rational function with a square root term is evaluated by using the substitution method or by using integration by parts. In this case, the substitution method can be used to simplify the integral and then solve it using basic integration techniques.

## 5. What is the significance of the definite integral of the given function?

The definite integral of the given function represents the exact area under the curve between the limits of integration. It can be interpreted as the total displacement of an object over a given time period or the total amount of work done by a variable force over a given distance. It also has applications in finding the average value of a function and calculating probabilities in statistics.

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