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anonymous
 5 years ago
integrate by partial fractions, not sure,. i want to integrate 1/((x^2*(x^215)) should i do partial fraction 1/((x^2*(x^215)) = A / x + B/x^2 + C/(xsqrt 15) + D/(x+sqrt 15) , or 1/((x^2*(x^215)) = A / x + B/x^2 + Cx+d / ( x^2  15)
anonymous
 5 years ago
integrate by partial fractions, not sure,. i want to integrate 1/((x^2*(x^215)) should i do partial fraction 1/((x^2*(x^215)) = A / x + B/x^2 + C/(xsqrt 15) + D/(x+sqrt 15) , or 1/((x^2*(x^215)) = A / x + B/x^2 + Cx+d / ( x^2  15)

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1oy... if you hit the enter key it spreads this out to where it is more readable.... just a thought

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the x^2  15 is reducible , but im not sure if youre supposed to use that theorem

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1the key on the keyboard that makes this happen ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you want me to use the enter key more often?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1yes, difference of squares that part and you should have 3 sets to figure out

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1radicals are fine.... right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0are they ? how come the other way works just as well?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1which other way can you decompose the fraction?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01/((x^2*(x^215)) = A / x + B/x^2 + (Cx+d) / ( x^2  15)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0technically the quadratic is reducible

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1it could be a fluke if it does work out....

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1or it could be a possibility for solutions that i am unaware of as yet :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no it isnt a fluke, in general one need not reduce a quadratic

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1in general you should, to account for multiples i think

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1if it aint got multiples, then you should be fine i spose

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes it usually makes it easier to reduce the quadratic, except in the case where there are radical linear factors

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it can have multiples, theres no problem as far as i can see

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1without doing a ton of them to generalize it, id have to throw it in the hmmmmm box lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0say integral ( 1 / (x(x+1)^2) = A/x + (Bx+C) / (x+1)^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its a little weird, might be easier if i wrote 1/( x ( x^2+2x+1)) = A/x + (Bx+c) / (x^2 + 2x + 1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sometimes it is difficult to factor a quadratic, right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok , i should get into more accurate terms. sometimes it is reducible over reals, not rationals

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1factoring a quadratic is fairly easy; might get complicated solutions, but farily easy to do

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes but doing partial fractions with irrational roots is tough

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1and i have heard that you can use the complex solutions as well with no ill effects on the decomp

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok , so youre so confident, try to do parfractions on 1 / ( x^2(x^215) , give it a go

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01/((x^2*(x^215)) = A / x + B/x^2 + C/(xsqrt 15) + D/(x+sqrt 15)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1confident? lol ; just stubborn

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it can even get stranger, this is an easy irrational solution problem, you can have say A/ ( x  sqrt 3/2 ) for instance

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this is why calculus in textbooks is generally contrived , or idealized problems. in the real world you have these irrational roots, etc

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0theorem time, one sec

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.11 a b c d  =  +  +  +  x^2(x^215) x x^2 xsqrt(15) x+sqrt(15) is the 'proper' set up right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Personally, I'd leave the last fraction's denominator without doing difference of two squares.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01 / [(xa)^n (x^2+bx+c)^n] = A1/(xa) + A2/ ( xa)^2 + ... An/(xa)^n + (B1x+C1)/ (x^2+bx+c) + (B2x + C2) / (x^2+bx+c)^2 + ... + (Bn x + Cn ) / (x^2+bx+c)^n

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so the question is, do the quadratics have to be irreducible, i dont think so

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.11 a b c d  =  +  +  +  x^2(x^215) x x^2 xsqrt(15) x+sqrt(15) a(x^2)(xsqrt(15))(x+sqrt(15))) b(x)(xsqrt(15))(x+sqrt(15))) c(x)( x^2)(x+sqrt(15)) d(x)(x^2)(xsqrt(15))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0MOST calculus books and websites say the quadratic is irreducible though

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0irreducible over the reals i imagine. wikipedia is better, they say irreducible over the field

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0whatever field you have in mine

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0For this particular problem: \[\frac{1}{y(y15)} \equiv \frac{1}{15(y15)}  \frac{1}{15y}\] \[\text{Let } y = x^2\] ...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0newton, be careful though

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if you are substituting a variable, you will need to do 2x dx = dy

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so you might want to just change it back. thats an identity

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i should write a book on calculus pitfalls, since i keep finding them :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0newtons, so just change y back to x^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry, in case it wasn't clear, it leads to: \[\frac{1}{x^2(x^215)} \equiv \frac{1}{15(x^215)} + \frac{1}{15x^2} \] I would integrate this.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0otherwise you have 2x dx = dy , dx = dy / ( 2x ) = dy / ( 2 sqrt y )

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0correct, i thought you were doing a change of variables

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.11 = a(x^2)(xsqrt(15))(x+sqrt(15))) + b(x)(xsqrt(15))(x+sqrt(15))) + c(x)( x^2)(x+sqrt(15)) +d(x)(x^2)(xsqrt(15)) when x = sqrt(15) we get 1 = c(sqrt(15))(15)(2sqrt(15)) 1 = c(15)(2)(15) 1 = c(225)(2) = c450 c = 1/450

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0amistre, you get medal for persistence :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but its much easier if we dont split the quadratic

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I was only doing that to show that you don't need a 1/x fraction and 1/x^2 , as x^2 is the only term. I wasn't integrating with a different variable.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1yay!! lol :) can I stop now lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait that doesnt work though

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I mixed up a minus and plus sign ¬_¬

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its incorrect to do 1 / ( x^2 (x^215 ) = A / x^2 + b / (x^215) , thats incorrect

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you wont be able to find A and B

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you used a simpler case though

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Multiply it out, plz.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01 = A(x^2  15) + B x^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0, 1 = Ax^2 + bx^2  15

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(the + the last fraction should be minus, as in the first time, typo)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01 = Ax^2 + Bx62  15 A

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but the book says it should be A/ x + B/x^2 + C x+d / (x^2  15)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well this approach works as well

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can do it if you want, but you'll find A = 0, so it doesn't really matter.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right, so come the theorem doesnt work in these websites

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0as far as repeated roots go

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for instance you didnt use CX+D, i think its because of that y = x^2 business,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There are only terms in x^2 in this case, it wouldn't always work. The general method is probably more ... general.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so my question, my original question, hmmmm, do the quadratics have to be irreducible in the general case

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok for instance, 1 / (( x2)^2 ( x^215))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01 / (( x2)^2 ( x^215)) = A / (x2) + b / (x2)^2 + (cx+d) / (x^2  15 ) ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'd say it's probably easier to leave it as the irreducible (without using difference of two squares) quadratic with Cx + d. I guess you could split up this factor, but it's unnecessary (as far as I know)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but its not technically irreducible

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i can even go so far as to do this

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01 / (( x2)^2 ( x^215)) = (Ax+b) / (x2)^2 + (cx+d) / (x^2  15 )

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so this whole irreducible business is unnecessary

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01 = a*x^315*a*x+b*x^215*b+c*x^34*c*x^2+4*c*x+d*x^24*d*x+4*d

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok im going to solve an easier one

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0int 1/ ((x2)(x^215))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01/ ((x2)(x^215)) = A/ (x2) + (Bx+c) / ( x^ 2  15)
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