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partial fraction madness, continuation

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method for finding \(b\) in \[\frac{5x^2+20x+6}{x(x+1)^2}=\frac{a}{x}+\frac{b}{x+1}+\frac{c}{(x+1)^2}\]
in terms of a and c and x ?
@mahmit2012 method with derivatives

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Other answers:

what's wrong with ordinary method of finding partial differential?
@mahmit2012 had a snap method, but i don't understand it
just a question: do wolfram has solution?
they have something ugly...
we get \(a=\frac{6}{1}=6\), and \(c=\frac{-9}{-1}=9\) but \[b=5+0-\frac{6}{(-1)^2}=-1\] step i don't get
oh the original question was an integral, btw.
something to do with a derivative but i don't see it
what's your method for B
is there link to where it was done in snappier way?
looks like the method was, take \[5x^2+20x+6\] divide by \(x\) get \[5x+20+\frac{x}{6}\] take the derivative get \[5-\frac{6}{x^2}\] and then evaluate at \(x=-1\) why this works is going to bug me
but is sure is snapp!
I think we can solve this making three equations....
i meant did you have your own way, satellite?
a slower... sane approach
i plugged in A and C.. expanded it all out... plugged in 1 for x and got B=-3
ooooooooooooooh!!!! multiply both sides by \((x+1)^2\)!!!!!
but that's not right i guess
wowee zowee
how cool is that? i love it
@alienbrain yes i wrote my method before solved \(6+B=5\) but it required some visualization
it is already solved, i was trying to understand @mahmit2012 approach
now i get it, maybe i can remember is @alienbrain do not multiply out, you will make an algebra mistake, it is a pain we know \(6+B=5\) and so \(B=-1\)
@mahmit2012 my less sophisticated method was to visualize the coefficient of \(x^2\) as \(6+B\)
@satellite73 are u sure your answer is correct.... because I got a=15 b=-10 and c=1
yeah i am sure
yes i see thanks i didn't understand that you multiplied on both sides
i think i would not want to use this method if i had to take the derivative repeatedly
this is the shortest method to get cofficiant in partial fraction problem >
back, open study lagged me out
yeah it was getting laggy hope we have worn this topic out sufficiently we have the answer and a plethora of methods
so 5=6 + B are you just combining like terms there
good thing you can't throw a math book at me:)
ok lets go back back back
we know \(A=6\)
first term on the left is \(A(x+1)^2=6(x+1)^2\) and we also have a term that looks like \(Bx(x+1)\)
whatever all the other mess is when you multiply out, the only way to get \(x^2\) is from the first term, which will give you \(6x^2\) and from the second term which will give you \(Bx^2\)
on the other hand we know we have to end up with \(5x^2\) that tells you \(6+B=5\) and so \(B=-1\)
ok, i got it... but the method seems to vary for each letter
i was hoping to avoid critical thinking and just be able to do it by a process :)
ok you can use one method if you like you can multiply this mess out and solve a 3 by 3 system, but i try to avoid that if at all possible because it is a ton of work try to look for values of \(x\) that will make everything zero obviously that will not work all the time when you have repeated factors but i find it much easier to use that method as far as i can once you have some of the constants, the others will be easier to find, even if you have a system of equations
if you have no repeated factors, the it always works
well, if they are linear
it is also very easy to make an algebra mistake when multiplying out and combining like terms, much few mistakes plugging in numbers and evaluating in any case if you were going to use "one method" you would have ended up with the equation \(A+B=5\) but think of how much easier if you know \(A=6\) already and we got \(A=6\) with almost no work
ok enough partial fractions for one night i wish @eliassaab was here to show another method it has to do with taking the limit as \(x\to \infty\) which for a rational function you can pretty much eyeball, and then determining what a constant is from that
eh.. integral of 6/x ?
oh nm

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