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The Fuck? Calculus Problem. Please help me.

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Gnarcoma
Posts: 437 - Karma: 5

Integrate (x^3+1)^(1/3)*x^5 with respect to x.

Could somebody show me how to do this problem? I can't even figure out where to start. 

You can copy+paste the function into this website to read it easier and check your answer: http://www.quickmath.com/webMathematica3/quickmath/page.jsp?s1=calculus&s2=integrate&s3=basic

Nov 8 2010 5:23PM
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2_fadez
Posts: 7780 - Karma: 4,180
so do you want the antiderivative or the derivative of that function

Nov 8 2010 5:36PM
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yuck
Posts: 8384 - Karma: 2,090
this is a bit over my head, probably gonna be seeign this next semester

but, www.khanacademny.org might help
Nov 8 2010 5:38PM
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Gnarcoma
Posts: 437 - Karma: 5
antiderivative.
Nov 8 2010 5:39PM
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notadoctor
Posts: 398 - Karma: 589
integral==antiderivative??
I would use substitution method
Nov 8 2010 5:42PM
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steevner
Posts: 4502 - Karma: 47
look up integration by parts.

Gimme some-of-that karma bitch!
Nov 8 2010 5:43PM
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antwone
Posts: 1352 - Karma: 157
use this (uv)'=u(v)'+(u)'v
Nov 8 2010 5:46PM
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MCIRONCOCK
Posts: 3094 - Karma: 30
Click for larger image
Nov 8 2010 5:51PM
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Gnarcoma
Posts: 437 - Karma: 5

That is if I wanted to find the derivative of this function.  I want to find the antiderivative.

To the people who have told me to do u-substitution:  I've tried that; I can't get it to work. 

Nov 8 2010 5:53PM
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J_jibbs
Posts: 10 - Karma: 15
Go to www.wolframalpha.com

It does calculus for you...

or track down an asian...
Nov 8 2010 5:56PM
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pmsnow
Posts: 992 - Karma: 87
wolframalpha.com
Nov 8 2010 5:56PM
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BootySpunk.
Posts: 8148 - Karma: 84
Lemme take a crack at it.
Nov 8 2010 5:57PM
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PhilboydStudge
Posts: 3431 - Karma: 206
Use integration by parts. Find u, du, dv, and v, and use u*v-integral of(v*du).

If that doesn't make sense, try wolframalpha.com. Best site in the world for math, shows you all the steps.
Nov 8 2010 5:58PM
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BootySpunk.
Posts: 8148 - Karma: 84
Probably this...I really don't feel like doing IBP right now because it fucking sucks dick.

But I may if I get bored
Nov 8 2010 6:00PM
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BootySpunk.
Posts: 8148 - Karma: 84
Also...I am pretty sure you messed up writing the problem? Unless you have like x raised to an x raised to an x.
Nov 8 2010 6:02PM
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LMPJibbin
Posts: 1607 - Karma: 17
4chan. NS doesnt do math..
Nov 8 2010 6:09PM
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Gnarcoma
Posts: 437 - Karma: 5
That is a multiplication sign in between the 1/3 and x^5.
Nov 8 2010 6:10PM
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i.ski.walls
Posts: 94 - Karma: 25
i just did the problem...so you can use u substitution. substitute u= (x^3 +1) and then your du= 3x^2 dx.

so then you end up with taking the antiderivative of u^(1/3)*[(x^5*du)/(3x^2)] and you can just take it from there. you substitute the x's again after canceling them out.

if you need help let me know!
Nov 8 2010 6:11PM
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BootySpunk.
Posts: 8148 - Karma: 84
Yah...that is too nuts for me to try but try to use IBP because it isn't a trig sub and U-sub won't work and it is isn't a partial fraction case. So integration by parts is the only thing left to try.
Nov 8 2010 6:11PM
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notadoctor
Posts: 398 - Karma: 589
I think he means the cube root of (x^3)+1 multiplied by x^5???
Nov 8 2010 6:12PM
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dubaiski
Posts: 961 - Karma: 37
i think it is 1/7x^7+1/6x^6
but maths at half past 3 in the morning while slightly drunk is never a good idea so check it
i think you cube root the braket, mulitiply it out and then it is a simple intergration.
Nov 8 2010 6:19PM
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Gnarcoma
Posts: 437 - Karma: 5
Ok, I finally figured it out.  Karma to I.ski.walls because she got me half the way there.
Nov 8 2010 6:19PM
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dubaiski
Posts: 961 - Karma: 37
and pm me if you have any questions, i think that the website did it a different way to e and mine is more simplified. May be the booze talking though. lol
peace
Nov 8 2010 6:26PM
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