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

- anonymous

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

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- anonymous

I just don't know how to set it up, what should u be?

- anonymous

say u = 1/(sqrt(x-2)
then you have integral of u/u^2-2. use partial fractions to solve.

- anonymous

sorry,u = 1/sqrt(x+2)

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- anonymous

lemme know how it works out.

- anonymous

ok thx I'll try it out

- anonymous

did you get du=-1/2(x+2)^3/2? or did I do something wrong

- anonymous

thats right, just checked it

- anonymous

ok great! . i'd actually not done it. I just looked at the problem and did some calculation in my head. sorry.

- anonymous

np, wait though not done yet

- anonymous

so next du=-1(x+2)/2(x+2)^1/2 right?

- anonymous

umm, you can solve it entirely in u and plug in the values of u, since this is a definite integral.

- anonymous

no need to deal with x at all

- anonymous

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

- anonymous

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.

- anonymous

ok

- anonymous

Ok, found it

- anonymous

hey!

- anonymous

Ok, so first
Let \(u = (x+2)^{1/2} \implies du = (1/2)(x+2)^{-1/2}dx\)
\[ \implies dx = 2(x+2)^{1/2} du\]

- anonymous

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

- anonymous

Whoops forgot the 2

- anonymous

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?

- anonymous

and does the integral change from -1 to 2, to 1 to 2, or is that a typo?

- anonymous

The integral changes because \(u = (x+2)^{1/2}\) so
when x = -1, u = 1 and when x = 2, u = 2

- anonymous

right right, forgot about that, when you said you forgot the 2, is it ]2u^2-2]

- anonymous

\(2[u^2 - 2]\)

- anonymous

ah right, that would make more sense, well thx again

- anonymous

Do you see how I arrived at \(2[u^2-2]\)?

- anonymous

Cause I kinda skipped some steps there.

- anonymous

well no, but I understand that it's the right equation to the integral

- anonymous

I can show the steps.

- anonymous

there are steps to it?

- anonymous

Yes certainly.

- anonymous

awesome

- anonymous

Since
\( u = (x+2)^{1/2} \implies x = u^2 -2\)
\(\implies dx = 2u\ du\)
\[\implies \frac{x}{(x+2)^{1/2}}dx = \frac{u^2-2}{u} *2u\ du\]
\[=2[u^2-2]\ du\]

- anonymous

awesome thanks a lot, I don't think we even learned that yet

- anonymous

do you always implicitly differentiate x? (x'=dx and not 1?)

- anonymous

I found that out nvm

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