Indefinite Integration of a Rational Expression

communitycoll
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Homework Statement


<Indefinite integral sign here>[r^2 -2r] / [r^3 - 3r^2 + 1]dr
or the second example in the "Substitution" section here:
http://people.clarkson.edu/~sfulton/ma132/parfrac.pdf

Homework Equations


nada.


The Attempt at a Solution


Nothing to really attempt, I just don't get what they do with the numerator (i.e., how it turns into a 1).
 
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hi communitycoll! :smile:
(have an integral: ∫ and try using the X2 button just above the Reply box :wink:)
communitycoll said:
… I just don't get what they do with the numerator (i.e., how it turns into a 1).

ah, nooo, the numerator isn't r2 - 2r, it's (r2 - 2r)dr !

(and of course that's 1 times dw :wink:)

the trick in substitution is that you always have to substitute the "d" part also! :smile:
 
If you read that attachment carefully, you would have seen this:
Solution: Here we notice that the numerator is the derivative of the denominator (to within a constant factor).
So they let
w = r3 - 3r2 + 1,
so
dw = 3r2 - 6r dr,
which is the same as
dw = 3(r2 - 2r) dr.

Do you see it now?
 
Okay then. Yeah, I understand now. Thanks.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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