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How to calculate the integral below:

$$ \int_{0}^{1}(ax^2+bx+c)^{-3/{2}}dx $$

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    What did you try? Where are you stuck?2011-11-29
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    For $a\ne0$, write $ax^2+bx+c={ a({ (x+{b\over 2a})^2+\underbrace{{c\over a} -{b^2\over 4a^2} }_D) }}$. There are three cases to consider now, depending on whether $D=0$, $D>0$, and $D<0$. Two of these require "inverse trigonometric substitutions". The other leads to a simple integral.2011-11-29
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    I think it would be a good idea if you did some "warm up" problems with $a$, $b$, and $c$ given particular values.2011-11-29
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    @Leaozinho - Okay, then complete the square and try a trigonometric substitution.2011-11-29
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    @Leaozinho - Not getting **what**? Again, try completing the square. If you don't know how to do that, then step away from calculus and learn how to complete the square. Then come back to the problem.2011-11-29
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    @Leaozinho - Okay, so once you complete the square, try a trigonometric substitution in the two cases that David Mitra pointed out in his comment above.2011-11-29

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