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Can someone help me find the integral of (x^p)ln(x)dx? thanks!
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Re: calculus help....
is p whatever exposant because I could help you. But I just cant seem to find a way to integrate it by variable switching if I dont know what is P -
Re: calculus help....
yesComment
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Re: calculus help....
from what im trying, the answer will be highly algebric when i'll try to find the primitiveComment
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Re: calculus help....
Well! I opened Maple (powerfull math software) and maple did it... It gave a very big stuff. Is the problem right.
Do you have msn, maybe I could show you what it gave meComment
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Re: calculus help....
We're doing the kids' homework now?
"Quiet, numbskulls, I'm broadcasting!" -Moe Howard, "Micro-Phonies" (1945)Comment
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Re: calculus help....
I can help you.. just a minuteComment
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Re: calculus help....
You have to do integration by parts.
You'll have to take the integral of one of those inner functions, and the derivative of the other.
Since ln(x) is easier to differentiate than to integrate and x^p is not that hard to do either way, I'd go with differentiating ln(x).
The answer will be [ln(x)*(1/x)*x^(p+1)] minus the integral of (1/x)*x^(p+1)*ln(x)
Then, that integral should be easy to evaluate by substitution of variables because you have ln(x) and its derivative, 1/x in the same function.
Let U = ln(x), then x^(p+1) is equal to (E^u)^(p+1)
If you need more help, let me know.
By the way, this is assuming p is a constant.
If you are in vector calculus and p is a variable, sorry but I can't help you yet [img]/images/graemlins/laugh.gif[/img]
EDIT--
oops, I goofed and wrote some ps where I needed x
guess what, x^(p+1) times 1/x is equal to x^p, no?Comment
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Re: calculus help....
[ QUOTE ]
We're doing the kids' homework now?
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Kids? lets see you try it gramps. [img]/images/graemlins/grin.gif[/img]Comment
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Re: calculus help....
Get offa my lawn, you punks! [img]/images/graemlins/laugh.gif[/img]
"Quiet, numbskulls, I'm broadcasting!" -Moe Howard, "Micro-Phonies" (1945)Comment
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Re: calculus help....
[ QUOTE ]
Get offa my lawn, you punks! [img]/images/graemlins/laugh.gif[/img]
[/ QUOTE ]
Nothing like chasing away the ever so adventurous and rebellious math nerds, who are always up to no good.Comment
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Re: calculus help....
[ QUOTE ]
You have to do integration by parts.
You'll have to take the integral of one of those inner functions, and the derivative of the other.
Since ln(x) is easier to differentiate than to integrate and x^p is not that hard to do either way, I'd go with differentiating ln(x).
The answer will be [ln(x)*(1/p)*x^(p+1)] minus the integral of (1/p)*x^(p+1)*ln(x)
Then, that integral should be easy to evaluate by substitution of variables because you have ln(x) and its derivative, 1/x in the same function.
Let U = ln(x), then x^(p+1) is equal to (E^u)^(p+1)
If you need more help, let me know.
By the way, this is assuming p is a constant.
If you are in vector calculus and p is a variable, sorry but I can't help you yet [img]/images/graemlins/laugh.gif[/img]
[/ QUOTE ]
Yea I was supposed to do it with integration by parts, but I got a really fookin ugly answer...
ln(x)[x^(p+1))/(p+1)] - [x^(p+1)/(p+1)^2]Comment
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Re: calculus help....
by the way, PLEASE look at my edited version of what I typed, I goofed at first haha.. I'm definitely not used to typing out this kind of crap, especially when I can't enter symbols for integrals and superscript and stuff...Comment
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Re: calculus help....
wait asecond I goofed it up again, I had it right the first time... it is 1/p times x to the p+1 damn itComment
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Re: calculus help....
look what maple gave for answer :
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