Evaluating this improper integral

[Sophdof1]

Hi everyone

I have spent a whole two days trying to integrate (x+2)^2/(sqrt(1-x^(-2)) with respect to x.

They want me to find the integral from 1 to 2 which I can do! It is the just the actual general integration method. I did find that substituting 1/x got me somewhere but I'm just lost.

 

[MarkFL]

Hello, and welcome to FMH!

I think I would begin by writing:

[MATH]I=\lim_{t\to1}\left(\int_t^2 \frac{x(x+2)^2}{\sqrt{x^2-1}}\,dx\right)[/MATH]
Next, I would consider integration by parts, where:

[MATH]u=(x+2)^2\implies du=2(x+2)\,dx[/MATH]
[MATH]dv=\frac{x}{\sqrt{x^2-1}}\,dx\implies v=\sqrt{x^2-1}[/MATH]
Hence:

[MATH]I=\lim_{t\to 1}\left(\left[(x+2)^2\sqrt{x^2-1}\right]_t^2-2\int_t^2 (x+2)\sqrt{x^2-1}\,dx\right)[/MATH]
Can you continue?

 

[Sophdof1]

H

Hello, and welcome to FMH!

I think I would begin by writing:

[MATH]I=\lim_{t\to1}\left(\int_t^2 \frac{x(x+2)^2}{\sqrt{x^2-1}}\,dx\right)[/MATH]
Next, I would consider integration by parts, where:

[MATH]u=(x+2)^2\implies du=2(x+2)\,dx[/MATH]
[MATH]dv=\frac{x}{\sqrt{x^2-1}}\,dx\implies v=\sqrt{x^2-1}[/MATH]
Hence:

[MATH]I=\lim_{t\to 1}\left(\left[(x+2)^2\sqrt{x^2-1}\right]_t^2-2\int_t^2 (x+2)\sqrt{x^2-1}\,dx\right)[/MATH]
Can you continue?

Hi, could you perhaps show me how integrate further ?

 

[Sophdof1]

Also how

H


Hi, could you perhaps show me how integrate further ?

Furthermore, does the limit exist because looking at what I have got so far it doesn’t look like it does.

Thanks

 

[MarkFL]

Can you post what you have so far?

 

[Steven G]

I just want to comment on the work that you did so far.
1st: It should be the limit as x-->1⁺, not 1^-. This is because if x <1 and goes to 2 it WILL pass 1 which you realized you did not want to happen
2nd: You have \(\displaystyle x=\frac{1}{u}\). Then \(\displaystyle x+2 =\frac{1}{u} + 2\)

At least doing it correctly you can see if this method will work. So will it work, that is can you solve this new integral???