roselin
evaluate indefinite integrals
integral squareroot x^4/ x^31 dx



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roselin
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what is it ?

roselin
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thanks

RadEn
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is the x^31 in the sqrt also or no ??

roselin
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yes, it is.

RadEn
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dw:1355676028137:dw

roselin
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okay,

RadEn
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now, use int by subs
let u=x^31
du = 3x^2 dx or
x^2 dx = 1/3 du
continue it.....

roselin
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1/3 intergral 1/ sqrt of u * du

RadEn
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miss () sign, how come this ?

roselin
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i took du as 3x^2

RadEn
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derivative of x^3 is 3x^2, right ?

roselin
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yes

RadEn
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so, why becomes () ?

roselin
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well, i wrote the problem as 1x^3 instead of x^3 1

RadEn
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lol, hahaha...

roselin
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hmm

RadEn
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so, the original problem is over x^31 or 1x^3 ?

roselin
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it is x^3 1

RadEn
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ok, therefore the integration can be
dw:1355676826761:dw

RadEn
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i just gave u +c :), for the rest let try it

roselin
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u= x^3 1 here

RadEn
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yes

RadEn
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because i have saw ur problem is .... over x^31, but experiment gave u 1x^3

roselin
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i know

roselin
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and i put it in the wolfram thing, got some different ans

RadEn
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what's the result of int above, dont use wolfram first

roselin
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(u) ^1/2 +1/ 1/2+1

RadEn
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where is the 1/3 ?

roselin
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(x^31)+c

roselin
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What do I do with that 1/3?

roselin
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can you please write down the whole problem and show how it works, if you do not mind then

roselin
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i am kind of confused

RadEn
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dw:1355677696473:dw

RadEn
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got it ?

roselin
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how did u get that 2u?

RadEn
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1/(1/2 + 1) = 1/(1/2) = 1 * 2/1 = 2 right ?

roselin
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yeah, sorry

roselin
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i have another problem, will you guide me through that one? i will try solving it and show it to you.

RadEn
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nopes... now, subtitute back that u=x^31
so, the answer be 2/3 (x^31)^(1/2) + c or
2/3*sqrt(x^31) + c

RadEn
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wolfram is same with the answer above :)

roselin
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k.

roselin
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okay

roselin
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thank you.

RadEn
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very welcome