January 12, 200719 yr I've spent 3 hours on this problem, and I'm guessing I'm missing something extremely simple. If any of you know how to integrate functions, I'd be obliged if you could give me a hand. ÃÆâÃâ¹Ã¢â¬Â Ãâë [x^3 * ÃÆâÃâ¹Ã¢â¬Â Ãâ¦Ã¡(x^2 + 1)] dx I'm totally stuck. And please don't respond unless you actually know how to integrate :P
January 12, 200719 yr We just started learning integration this week =P. But isn"t there supposed to be a closed interval, like with 2 numbers on the top and bottom of the integral symbol? I guess that must only be with the ones we are doing. And please dont respond unless you actually know how to integrate :P I do know how to integrate, just not thise one :anxious: Sorry if you consider this spam Runescape, what was the point of that, cant you read? Last.fm: http://www.last.fm/user/Aaronm14/MY FAVORITE BAND:http://profile.myspace.com/index.cfm?fu ... d=64310717And the bible is the big book of lies, call me a racist if you must.
January 12, 200719 yr Author But isn"t there supposed to be a closed interval, like with 2 numbers on the top and bottom of the integral symbol? I guess that must only be with the ones we are doing. That's a definite integral, which returns a value. My question is an indefinite integral, which returns a function.
January 12, 200719 yr X^5 +1 that what i have so far.I'll get back to you when I have a different anwser
January 12, 200719 yr Author X^5 +1 that what i have so far.I'll get back to you when I have a different anwser You are extremely far off, to the point that I don't think you know what integration is.
January 13, 200719 yr X^5 +1 that what i have so far.I'll get back to you when I have a different anwser You are extremely far off, to the point that I don't think you know what integration is. Meh, get on MSN =P. I'm using U-substitution right now, but uh I'm stuck on a step. Might have to review me rules a bit more, but to be honest, so far I really only know how to do just simple anti-derivs and u-subs :(
January 13, 200719 yr you'll have to use integration by parts, but i gave up maths and i cba to work it out for you. babelfish - level 180 60th placestrongguy - level 173 69th place
January 13, 200719 yr I tried integration by parts with u = x^2 and dv = xÃÆâÃâ¹Ã¢â¬Â Ãâ¦Ã¡(x^2 + 1) to no avail. Though I'm incompetent with integration so perhaps you can do better :P
January 13, 200719 yr just curious, what class is this for? It's dealing with integrals so it is some form of Calculus.
January 13, 200719 yr Well I think I have it figured out by I have no time to finish it really :P It's integration by parts, that I'm fairly sure of. u = (x^2+1)^(1/2) v' = x^3 u' = x(x^2+1)^(-1/2) v = (x^4)/4 Integral uv' dx = [ (x^2+1)^(1/2) * (x^4)/4 ] - [(1/4) Integral x(x^2+1)^(-1/2) * (x^4)/4 ] Then you have to do integration by parts again on [(1/4) Integral x(x^2+1)^(-1/2) * (x^4)/4 ] Does that help? The answer SHOULD be (1/15)*(x^2+1)^(3/2)*((3x^2)-2) but I didn't get it through going through the steps. There is a handy program which can do pretty much any integral for you without showing the steps (obviously) and it can be found at integrals.wolfram.com By The_Jeppoz :wink:
January 13, 200719 yr Author The wolfram integrator puts it in extremely odd forms, I've found. Anyways, azn, mxm and punky, those were both extremely helpful, thanks a bunch!
January 13, 200719 yr dont you have to use the product rule as your multiplying two functions... neway thts what we call it where im from... unfortunately im to lazy to work it out....
January 13, 200719 yr Author dont you have to use the product rule as your multiplying two functions... neway thts what we call it where im from... unfortunately im to lazy to work it out.... Product rule is differentiation. The integral method of product rule is called integration by parts.
Create an account or sign in to comment