I'm going to be pathetic and ask for homework help...

[Random Variable]

In the process of trying to find the length of a curve, I'm left with the integral (4+4x^2+1/x^2)^(1/2)dx. Maple's answer is [((4x^2+4x^4+1)/(x^2))^(1/2)*x(x^2+ln(x))]/(1+2x^2), which evaluted at the proper limits gives me the correct answer. But what the heck did Maple do?

 

[acemcmac]

Is the answer "women are the root of all evil" for 500 alex?

 

[fbrdphreak]

Who cares?

*copy/paste maple answer, turn in HW*

Moving on...

 

[Mayfriday0529]

Can you plagiarise math?

 

[Papagayo]

she replaced "t" with "x"..

 

[sao123]

integration by calculator...
dig out the TI89.

 

[johnjbruin]

Isnt that just integrating ((1/t)+2t)dt ?

 

[hevnsnt]

"This is the sound a doggy makes"

 

[akubi]

Originally posted by: johnjbruin
Isnt that just integrating ((1/t)+2t)dt ?

it is.

 

[Kyteland]

Originally posted by: johnjbruin
Isnt that just integrating ((1/t)+2t)dt ?

Yes. sqrt(4t^2+4+1/t^2)
=sqrt((2t^2+1)^2)/t^2)
=(2t^2+1)/t
=2t+1/t

Also, [((4x^2+4x^4+1)/(x^2))^(1/2)*x(x^2+ln(x))]/(1+2x^2) reduces to
[((4x^2+4x^4+1)/(x^2))^(1/2)*x(x^2+ln(x))]/(1+2x^2)
=[(2x^2+1)/x*x(x^2+ln(x))]/(1+2x^2)
=x^2+ln(x)

...which matches up with the integral of 2t+1/t.

 

[icelazer]

Or, you can be smart about it, think about multilpying the top and bottom by sqrt(x^2)...factor, simplify, etc etc. x^2 + ln(x) + C

 

[Random Variable]

Originally posted by: johnjbruin
Isnt that just integrating ((1/t)+2t)dt ?

I can't believe I didn't think about adding up the terms inside of the square root. Man, I am stupid.