Partial fractions, chain rule

  • Thread starter EvLer
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  • #1
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Hi, I have 2 questions:

1. partial fractions:
if I have following integral: Itegral[(1-2x^2)/(x - x^3)]dx;
my question is do I break down the denominator to x(1-x^2) or do I go further:
x(1-x)(1+x); this way it becomes more complicated;

2. chain rule:
how does chain rule work in this case: f(x)g(x)h(x)?
is it something like f'(x)[g(x)h(x)] + g'(x)[f(x)h(x)] + h'(x)[f(x)g(x)] ??? or how would I differentiate this?

Thanks in advance.
 

Answers and Replies

  • #2
StatusX
Homework Helper
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1. You need enough unknowns to be able to equate the three coefficients for each power of x in the numerator (1,0,-2). So you'll need to break it up into all three factors. In some cases (not this one) the two factors alone would work, but only if the numerator happens to be just right. It would be like finding the point where two lines intersect in 3D; you can only find it if the lines happen to intersect.

2. Think of it as (fgh)' = f'(gh) + f(gh)'. Your guess is right.
 
  • #3
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In first 1-2x^2=1-x^2-x^2 is very helpful. Then you have 2 very easy integrals
 
  • #4
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Aaaah, i see. Thanks for help.
 
  • #5
OlderDan
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#2 is not the chain rule. That is the product rule. It does work as you have it
 
  • #6
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OlderDan said:
#2 is not the chain rule. That is the product rule. It does work as you have it
Right, sorry, mixed up the names :redface:
Chain rule is differentiating what I call "embedded functions".
 
  • #7
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EvLer said:
Right, sorry, mixed up the names :redface:
Chain rule is differentiating what I call "embedded functions".

Composite functions is their real name, btw.
 
  • #8
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*mental note to self* TERMINOLOGY!
I knew it, but did not recall on the spot, thanks everyone!
 

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