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evaluate indefinite integrals integral squareroot x^4/ x^3-1 dx

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what is it ?
is the x^3-1 in the sqrt also or no ??

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Other answers:

yes, it is.
now, use int by subs let u=x^3-1 du = 3x^2 dx or x^2 dx = 1/3 du continue it.....
-1/3 intergral 1/ sqrt of u * du
miss (-) sign, how come this ?
i took du as -3x^2
derivative of x^3 is 3x^2, right ?
so, why becomes (-) ?
well, i wrote the problem as 1-x^3 instead of x^3 -1
lol, hahaha...
so, the original problem is over x^3-1 or 1-x^3 ?
it is x^3 -1
ok, therefore the integration can be |dw:1355676826761:dw|
i just gave u +c :), for the rest let try it
u= x^3 -1 here
because i have saw ur problem is .... over x^3-1, but experiment gave u 1-x^3
i know
and i put it in the wolfram thing, got some different ans
what's the result of int above, dont use wolfram first
(u) ^-1/2 +1/ -1/2+1
where is the 1/3 ?
What do I do with that 1/3?
can you please write down the whole problem and show how it works, if you do not mind then
i am kind of confused
got it ?
how did u get that 2u?
1/(-1/2 + 1) = 1/(1/2) = 1 * 2/1 = 2 right ?
yeah, sorry
i have another problem, will you guide me through that one? i will try solving it and show it to you.
nopes... now, subtitute back that u=x^3-1 so, the answer be 2/3 (x^3-1)^(1/2) + c or 2/3*sqrt(x^3-1) + c
wolfram is same with the answer above :)
thank you.
very welcome

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