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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)

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

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

ok

Ok, found it

hey!

So then
\[\int_{-1}^2\frac{x}{(x+2)^{1/2}}dx = \int_1^2[u^2-2]du\]

Whoops forgot the 2

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

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