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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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0either way worked out

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[1/15\int\limits_{}^{}(x ^{2}(x ^{2}15))dx/(x ^{2}(x ^{2}15))\]\[1/15[\int\limits_{}^{}dx/(x ^{2}15)\int\limits_{}^{}dx/x ^{2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes that works, but i want to use partial fractions

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^215)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats some trick you got there :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0partial fraction will be avoided when you have better choices...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the tricks will help u and its very time saving...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i guess was wondering, is it ok to have , a quadratic that is not reducible , ok let me state this correctly

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how did you know to seperate them like that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0only practice..........

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i can make it harder, one sec

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0int (1/ (x^3 (x^215))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok thats a good partial fractions problem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0here you can use partial fractions because no other way............

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, now we have two options it would seem, my question really boils down to this quadratic

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the quadratic is not irreducible,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0integrate 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 /x^3 +D/(xsqrt 15) + e/(x+sqrt 15) , or 1/((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.0here u can use the one which i gave first to reduce it . it would give you \[1/15[\int\limits_{}^{}dx/(x(x ^{2}15)\int\limits_{}^{}dx/x ^{3}]\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0my question was, both approaches are legal?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0does that work, im not sure

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh ok, i thought the quadratic MUST be irreducible

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but when u try to find cx+d/(x^215) then may be again u have to do partial fractions...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0are you sure this is correct 1/15[∫dx/(x(x2−15)−∫dx/x3], i did it and got a different answer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0[ x^3  x(x^215) ] / (x^4 (x^2  15)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im trying to reverse your result, checking if its equivlanet

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0((x^2)(x^215))/(x^3(x^215))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0again there is another way to do it just substitute x^2=y.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats not going to work, since 2x dx = dy

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it is again more easier ....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh you mean for the first problem ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0your integration is xdx/(x^4(x^215)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for the second one...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0try it... it surely works..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh , but its not simple

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lets try it for the first integral

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its not for the first one.... its only applicable for the second one....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0these some shortcut to approach....

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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do you agree to that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0have u understood properly?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, i just want to make sure i understand your subsitution

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dx/(x^3(x^215) gives you dy/(2y^2(y15))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i get dy / ( 2 y^(3/2) (y15)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0y=x^2 dy=2xdx dx/(x^3(x^215))= 2xdx/(2x^4(x^215)) = dy/(2y^2(y15))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you have to solve for dx

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0check it , thats correct...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.02x dx = dy , so dx = dy/(2x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok...let me to make u clear...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh, you multiplied top and bottom by 2x, one sec

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0be fast i have to go .... but its nice to solve maths with you?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sure we will talk again

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats the magic of maths...........

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0my way probably works, i have to work on it :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i live in usa, near new york

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0in which class are u?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.02x dx = dy , so dx = dy/(2x) , so dx = dy/ ( 2 sqrt y)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but it comes out the same

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0since x^3 = sqrt y ^3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0are u a student in college or school?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0very nice, both ways work

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i like your way though

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it never occured to me you can multiply top and bottom by x
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