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Does this indefinite integral converge for anyone? Also, if you are having trouble with the integral, take a look at the derivatives of inverse hyperbolic functions. --[[User:Jmason|John Mason]] | Does this indefinite integral converge for anyone? Also, if you are having trouble with the integral, take a look at the derivatives of inverse hyperbolic functions. --[[User:Jmason|John Mason]] | ||
| − | It does not converge for me. I used direct comparison to test whether it converges or not. I started by comparing <math>\sqrt{x^4-1}</math> and <math>\sqrt{x^4}</math> It was easy from there. | + | It does not converge for me. I used direct comparison to test whether it converges or not. I started by comparing <math>\sqrt{x^4-1}</math> and <math>\sqrt{x^4}</math> It was easy from there. --Josh Visigothsandwich |
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| + | Thanks. I'm not really sure that I ''needed'' to use some sort of comparison to show it didn't converge, as it did integrate nicely, but its good to have a second opinion, be that mathematical or otherwise. --[[User:Jmason|John Mason]] | ||
Revision as of 08:12, 26 October 2008
8.8 #54
Does this indefinite integral converge for anyone? Also, if you are having trouble with the integral, take a look at the derivatives of inverse hyperbolic functions. --John Mason
It does not converge for me. I used direct comparison to test whether it converges or not. I started by comparing $ \sqrt{x^4-1} $ and $ \sqrt{x^4} $ It was easy from there. --Josh Visigothsandwich
Thanks. I'm not really sure that I needed to use some sort of comparison to show it didn't converge, as it did integrate nicely, but its good to have a second opinion, be that mathematical or otherwise. --John Mason
