#1
I've got my AP Calculus A/B final tomorrow, and I'm leafing through the book, looking for difficult exercises. However, I have no freakin' idea how to find the indefinite integral of the following function:

Integral of ((x^2)-1)/((2x-1)^.5) dx

Can anybody who's decent at calc help? I tried u substition, making u = 2x -1, and then du = 2 dx, but I can't get it work. Thanks guys.
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#2
i don't think u substitution is gonna work. try to use radical substitution or partial fractions if you have gotten that far.
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#3
Quote by psychokiller99
i don't think u substitution is gonna work. try to use radical substitution or partial fractions if you have gotten that far.


Blargh, we haven't gotten there yet. It's in the u substitution section anyway, so, I'm completely lost. Thanks anyway though.
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#4
I don't know about that but I can tell you that 2+2=5.
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#5
I took BC and we didnt' have a final!

Anyway, the answer is somewhat complicated (TI-89 FTMFW). Try partial fractions.
#6
We haven't done partial fractions yet :/
Maybe it was a misprint or something?
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#7
could be. i don't see any way to do that by method of u-sub. try seeing if you can find a way to simplify it. i doubt you could, but maybe i'm missing something. if not, then idk. calculator or look-up table FTW.

just use any method you know. it's not like you won't be able to do that on the AP exam. they won't have a u substitution only section.
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#9
((x^2)-1)^2
----------------
2

take Switch the / with X and take the (2x-1)^5 and make it (2x-1)^-5

then u'll have ((x^2)-1) (2x-1)^-5


Edit : you know what forget it ,i've been taking this stuff in arabic so it'll be a little bit difficult for me to explain

Yeah ,it's not correct ,sorry
Wolv
Last edited by Thedarkwolf at Jun 9, 2008,
#10
try expanding it into ((x^2)/(2x-1)^.5)-(1/(2x-1)^.5)...the second half of the intergral is easy, i just dont know what the first one is. my calc is rusty


Quote by Thedarkwolf
((x^2)-1)^2
----------------
2


that leaves out the square root of 2x-1 in the original equation
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#11
i don't why but i thought it said catctus help and when i entered the thread i was like wtf is this

Oh yeah definetly expand it like that guy ^
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#12
u = 2x - 1 ==> x = (u + 1)/2

Plug into the original equation to get:

integral: [((u+1)/2)^2 - 1] / sqrt(u) du ==> integral: [(u^2)/4 + u/2 - 3/4] / sqrt(u) du

Simplify and integrate. Plug in 2x -1 for u when you are done.
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Last edited by darkstar2466 at Jun 9, 2008,
#13
Do it with the integration by parts formula, just make the bottom bracket's index a negative number and bring it to the top (if you know what i mean)

u*dv = uv - (integral)v*du

EDIT: ignore me, gandalf above seems to know his shiz
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#14
Quote by MakeItY0urs
try expanding it into ((x^2)/(2x-1)^.5)-(1/(2x-1)^.5)...the second half of the intergral is easy, i just dont know what the first one is. my calc is rusty


that leaves out the square root of 2x-1 in the original equation


My original intention was to cross the (2x-1) out from the equation as being a derivant from the ((X^2)-1) but it's not

so i think there has to be something im not thinking of
Wolv
#15
Quote by GarlicBread
Do it with the integration by parts formula, just make the bottom bracket's index a negative number and bring it to the top (if you know what i mean)

u*dv = uv - (integral)v*du


IBP takes more work than needed for this problem. The method I suggested goes pretty quick.
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#16
Quote by darkstar2466
u = 2x - 1 ==> x = (u + 1)/2

Plug into the original equation to get:

integral: [((u+1)/1)^2 - 1] / sqrt(u) du ==> integral: [(u^2)/4 + u/2 - 3/4] / sqrt(u) du

Simplify and integrate. Plug in 2x -1 for u when you are done.


Ahh, I see. Wow, I'm screwed if that's on the test

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#17
Quote by g.siddarth
Ahh, I see. Wow, I'm screwed if that's on the test

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Take comfort in the facts that:

1) There definitely WILL be "nice" answers to your questions.
2) You are not being given a system of three or four second order partial differential equations. (Shit gets worse over the years...)
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#18
wtf... i didnt even know you could do **** like that and i just passed calc 3...
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#19
move the (2x-1)^5 to the top and make the power negative, then use integration by parts;
integral f g' = f g - integral f' g
make one term f and one g'... whichever ones easier.. do that til it works
#20
Quote by Pricycle
move the (2x-1)^5 to the top and make the power negative, then use integration by parts;
integral f g' = f g - integral f' g
make one term f and one g'... whichever ones easier.. do that til it works


Having fun bumping old threads there? Especially ones that were answered the day they started?