Anyone want to help with the integral of x/(x+2)^1/2 from -1 to 2?

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Anyone want to help with the integral of x/(x+2)^1/2 from -1 to 2?

Mathematics
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I just don't know how to set it up, what should u be?
say u = 1/(sqrt(x-2) then you have integral of u/u^2-2. use partial fractions to solve.
sorry,u = 1/sqrt(x+2)

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lemme know how it works out.
ok thx I'll try it out
did you get du=-1/2(x+2)^3/2? or did I do something wrong
thats right, just checked it
ok great! . i'd actually not done it. I just looked at the problem and did some calculation in my head. sorry.
np, wait though not done yet
so next du=-1(x+2)/2(x+2)^1/2 right?
umm, you can solve it entirely in u and plug in the values of u, since this is a definite integral.
no need to deal with x at all
wait how come? I thought we had to make du look like the remaining terms in the integral, sorry I've just learned integrals today
Okay. let me get back to you with this. I will take out my pen and paper and write it down. Do ask someone else too.
ok
Ok, found it
hey!
Ok, so first Let \(u = (x+2)^{1/2} \implies du = (1/2)(x+2)^{-1/2}dx\) \[ \implies dx = 2(x+2)^{1/2} du\]
So then \[\int_{-1}^2\frac{x}{(x+2)^{1/2}}dx = \int_1^2[u^2-2]du\]
Whoops forgot the 2
wow, ok let me try that out thanks, do u know of any rules to find these? or to you just have to guess the u sometimes?
and does the integral change from -1 to 2, to 1 to 2, or is that a typo?
The integral changes because \(u = (x+2)^{1/2}\) so when x = -1, u = 1 and when x = 2, u = 2
right right, forgot about that, when you said you forgot the 2, is it ]2u^2-2]
\(2[u^2 - 2]\)
ah right, that would make more sense, well thx again
Do you see how I arrived at \(2[u^2-2]\)?
Cause I kinda skipped some steps there.
well no, but I understand that it's the right equation to the integral
I can show the steps.
there are steps to it?
Yes certainly.
awesome
Since \( u = (x+2)^{1/2} \implies x = u^2 -2\) \(\implies dx = 2u\ du\) \[\implies \frac{x}{(x+2)^{1/2}}dx = \frac{u^2-2}{u} *2u\ du\] \[=2[u^2-2]\ du\]
awesome thanks a lot, I don't think we even learned that yet
do you always implicitly differentiate x? (x'=dx and not 1?)
I found that out nvm

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