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Have you tried completing the square inside the square root first?
This might end up being a trig sub.
i will be back if you have questions

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Other answers:

yeah i tried it.. but that didnt worked !
u can take (x-1) as u and the rest term as v and integrate by parts
@ikram002p please help !
addy :o is it like this ? \(\int (x-1) \sqrt (1-x-x^2)\)
yeah exactly!
The trig sub seems to work Not saying there aren't other ways though
@aayushi.somani may I ask how far you got with that method?
to what integral have you transformed it?
i could not figure out anything
can you tell me what substitutions you performed ?
some people like to do two here i just like one
When you say you could not figure out anything does that mean you did not do anything with the trig sub even though you said you did?
I'm confused
i tried getting a trig sub but unfortunatly i couldnt.. thats what it is.. @ikram002p could u figure out?
did you complete the square inside the square root part?
if i could have done it then y would i ask this ques..??? @myininaya i asked so that anyone could help me..
Oh I'm sorry I didn't know you couldn't do the algebra either
So by nothing means absolutely nothing I get it
from what i did m getting a wrong answer so consider i did nothing and now please help me from the very begnning..
\[ax^2+bx+c=a(x^2+\frac{b}{a}x)+c=a(x^2+\frac{b}{a}x+(\frac{b}{a})^2)+c-a(\frac{b}{a})^2 \\ =a(x+\frac{b}{a})^2+c-a \frac{b^2}{a^2}=a(x+\frac{b}{a})^2+c-\frac{b^2}{a}\] I hope this helps you complete the square Please follow these steps I prefer you ask me questions on how to do this than just using the end result there
|dw:1408601698600:dw| i dont know if this work next to use tring mmm sin maybe ?
oh mine what is all of that?
u mean i gotto Rationalise? @ikram002p
You don't have to rationalize...And tht wouldn't be the way to rationalize the top anyhow
:P ok we dont need conjucate xD just hmm 1-x-^2=(5/2-(x+1/2)^2 >.<
got it lol sorry :P
i got the answer.. anyways thanks (:
:o how did u get it ?
I just made a small mistake.. I rechecked it.. and i could figure out the mistake! thankyou so much.. @ikram002p
:* np girl
I thought you didn't do any of it...
I want to go back to my completing square thingy... (i made some mistakes) \[ax^2+bx+c=a(x^2+\frac{b}{a}x)+c=a(x^2+\frac{b}{a}x+(\frac{b}{2a})^2)+c-a(\frac{b}{2a})^2 \\ =a(x+\frac{b}{2a})^2+c-a \frac{b^2}{4a^2}=a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a} \]

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