How do you take the antiderivative of (2x-1)^-1/2???

- anonymous

How do you take the antiderivative of (2x-1)^-1/2???

- chestercat

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- anonymous

Is that (2x-1)^(-1/2) or (2x-1)^-1 which is divided by 2?

- anonymous

(2x-1)^(-1/2)

- anonymous

O.k. Then you would let u = 2x-1
Then du = 2 dx
So your integral:\[\int\limits \frac{1}{\sqrt{2x+1}}dx\]
becomes

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## More answers

- anonymous

1/2 times \[\int\limits \frac{2dx}{\sqrt{2x+1}}\]

- anonymous

\[\int\limits \frac{du}{\sqrt{u}}=\]

- anonymous

\[\int\limits u ^{-1/2}du=\]

- anonymous

\[\frac{u ^{-1/2 + 1}}{-1/2+1}+C\]

- anonymous

\[\frac{(2x+1) ^{1/2}}{1/2}+C\]

- anonymous

and dividing by 1/2 is the same as multiplying by the reciprocal 2/1

- anonymous

What about x/(2x-1)^(1/2) with an upper of 5 and a lower of 1. My calculator is giving me a correcct answer of 5.333333. But I can work the problem out to be the same

- anonymous

sorry, I "can't" work the problem out the same

- anonymous

o.k. well, you would want to plug 5 into \[2*\sqrt{2x+1}\] and then subtract what you get when you plug 1 into it.

- anonymous

that would be 2 times the result of
\[\sqrt{2(5)+1} - \sqrt{2(1)+1} = \sqrt{11}-\sqrt{3}\]

- anonymous

which turns out to be about 3.17

- anonymous

Oh! I didn't notice you had an "x" in the numerator.

- anonymous

You would still use the same substitution

- anonymous

but the x in the numerator can be rewritten in terms of "u"

- anonymous

right. x= (1+u)/2

- anonymous

since u = 2x+1
u-1 = 2x
and so you get that
x = (.5)*(u - 1)

- anonymous

so dividing that by \[\sqrt{u} = u^{1/2}\]
you can use exponent rules to get

- anonymous

\[\frac{.5*(u-1)}{u^{1/2}} \]

- anonymous

which boils down to
\[.5* (u^{1-1/2}-u^{-1/2})\]

- anonymous

which is\[.5* (u^{1/2}-u^{-1/2})\]

- anonymous

and you can use the power rule for integrals to antidifferentiate

- anonymous

\[.5* (u^{3/2}-u^{1/2})\]

- anonymous

or
\[.5* ((2x+1)^{3/2}-(2x+1)^{1/2})\]

- anonymous

does that make sense?

- anonymous

Yes, but when I plug in the upper and lower I'm still not getting the correct answer.......

- anonymous

let me check

- anonymous

The (2x+1) is suppose to be (2x-1)

- anonymous

right...it's 14.8511...

- anonymous

Oh! :)
Well, I can guarantee you if you use 2x-1 instead it should work
but I'll check

- anonymous

My graphing calculator is giving me 5.333333

- anonymous

\[9^{3/2}-9^{1/2}-(1^{3/2}-1^{1/2})\]

- anonymous

is \[3\sqrt{3}-3\]

- anonymous

Do you have a graphing cal on you?

- anonymous

Actually, you have to go back and change the substitution also so it is more work than this

- anonymous

if u = 2x-1 then x=.5*(u+1)

- anonymous

Thanks for your help. I'm in FLorida and it's about 1am. I need to hit the hay. Thanks again......

- anonymous

Yeah, I did it by hand and it works out.
Just use that substitution

- anonymous

and also multiply by the reciprocal of the new exponent you get when you integrate. ;)

- anonymous

Will do. Thanks!

- anonymous

Your welcome...sorry I'm new to typing this out. It's a bit distracting ;) but I'll get used to it. Night!

- anonymous

Yeah, it's a pain in the retrice Laters

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