anonymous
  • anonymous
integrate by partial fractions, not sure,. i want to integrate 1/((x^2*(x^2-15)) should i do partial fraction 1/((x^2*(x^2-15)) = A / x + B/x^2 + C/(x-sqrt 15) + D/(x+sqrt 15) , or 1/((x^2*(x^2-15)) = A / x + B/x^2 + Cx+d / ( x^2 - 15)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
no
anonymous
  • anonymous
either way worked out
anonymous
  • anonymous
i checked my program

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anonymous
  • anonymous
\[1/15\int\limits_{}^{}(x ^{2}-(x ^{2}-15))dx/(x ^{2}(x ^{2}-15))\]\[1/15[\int\limits_{}^{}dx/(x ^{2}-15)-\int\limits_{}^{}dx/x ^{2}\]
anonymous
  • anonymous
now go ahead.
anonymous
  • anonymous
yes that works, but i want to use partial fractions
anonymous
  • anonymous
1/(x^2(x^2-15)) = A/x + B/x^2 + (cx+d)/(x^2-15)
anonymous
  • anonymous
thats some trick you got there :)
anonymous
  • anonymous
partial fraction will be avoided when you have better choices...
anonymous
  • anonymous
the tricks will help u and its very time saving...
anonymous
  • anonymous
i guess was wondering, is it ok to have , a quadratic that is not reducible , ok let me state this correctly
anonymous
  • anonymous
how did you know to seperate them like that
anonymous
  • anonymous
only practice..........
anonymous
  • anonymous
i can make it harder, one sec
anonymous
  • anonymous
int (1/ (x^3 (x^2-15))
anonymous
  • anonymous
ok thats a good partial fractions problem
anonymous
  • anonymous
yea...
anonymous
  • anonymous
here you can use partial fractions because no other way............
anonymous
  • anonymous
ok, now we have two options it would seem, my question really boils down to this quadratic
anonymous
  • anonymous
the quadratic is not irreducible,
anonymous
  • anonymous
integrate by partial fractions, not sure,. i want to integrate 1/((x^2*(x^2-15)) should i do partial fraction 1/((x^2*(x^2-15)) = A / x + B/x^2 + C /x^3 +D/(x-sqrt 15) + e/(x+sqrt 15) , or 1/((x^2*(x^2-15)) = A / x + B/x^2 + Cx+d / ( x^2 - 15)
anonymous
  • anonymous
here 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
  • anonymous
my question was, both approaches are legal?
anonymous
  • anonymous
does that work, im not sure
anonymous
  • anonymous
both are correct.
anonymous
  • anonymous
oh ok, i thought the quadratic MUST be irreducible
anonymous
  • anonymous
but when u try to find cx+d/(x^2-15) then may be again u have to do partial fractions...
anonymous
  • anonymous
are you sure this is correct 1/15[∫dx/(x(x2−15)−∫dx/x3], i did it and got a different answer
anonymous
  • anonymous
what did u get?
anonymous
  • anonymous
[ x^3 - x(x^2-15) ] / (x^4 (x^2 - 15)
anonymous
  • anonymous
no...
anonymous
  • anonymous
im trying to reverse your result, checking if its equivlanet
anonymous
  • anonymous
((x^2)-(x^2-15))/(x^3(x^2-15))
anonymous
  • anonymous
oh it works
anonymous
  • anonymous
ok
anonymous
  • anonymous
my math was off,
anonymous
  • anonymous
again there is another way to do it just substitute x^2=y.
anonymous
  • anonymous
thats not going to work, since 2x dx = dy
anonymous
  • anonymous
it is again more easier ....
anonymous
  • anonymous
then dx = dy / (2x )
anonymous
  • anonymous
oh you mean for the first problem ?
anonymous
  • anonymous
your integration is xdx/(x^4(x^2-15)
anonymous
  • anonymous
for the second one...
anonymous
  • anonymous
try it... it surely works..
anonymous
  • anonymous
oh , but its not simple
anonymous
  • anonymous
lets try it for the first integral
anonymous
  • anonymous
oh ok , so s
anonymous
  • anonymous
its not for the first one.... its only applicable for the second one....
anonymous
  • anonymous
these some shortcut to approach....
anonymous
  • anonymous
i have 2 go now.....
anonymous
  • anonymous
dx = dy / ( 2 sqrt y )
anonymous
  • anonymous
do you agree to that?
anonymous
  • anonymous
have u understood properly?
anonymous
  • anonymous
yes, i just want to make sure i understand your subsitution
anonymous
  • anonymous
dx/(x^3(x^2-15) gives you dy/(2y^2(y-15))
anonymous
  • anonymous
i get dy / ( 2 y^(3/2) (y-15)
anonymous
  • anonymous
y=x^2 dy=2xdx dx/(x^3(x^2-15))= 2xdx/(2x^4(x^2-15)) = dy/(2y^2(y-15))
anonymous
  • anonymous
got it?
anonymous
  • anonymous
thats wrong
anonymous
  • anonymous
you have to solve for dx
anonymous
  • anonymous
what?
anonymous
  • anonymous
check it , thats correct...
anonymous
  • anonymous
2x dx = dy , so dx = dy/(2x)
anonymous
  • anonymous
ok...let me to make u clear...
anonymous
  • anonymous
oh, you multiplied top and bottom by 2x, one sec
anonymous
  • anonymous
yesssssssssss
anonymous
  • anonymous
be fast i have to go .... but its nice to solve maths with you?
anonymous
  • anonymous
very cleverl
anonymous
  • anonymous
thanks , bye
anonymous
  • anonymous
sure we will talk again
anonymous
  • anonymous
thats the magic of maths...........
anonymous
  • anonymous
my way probably works, i have to work on it :)
anonymous
  • anonymous
my substitution
anonymous
  • anonymous
what do u do?
anonymous
  • anonymous
i solve for dx =
anonymous
  • anonymous
from where r u?
anonymous
  • anonymous
i live in usa, near new york
anonymous
  • anonymous
in which class are u?
anonymous
  • anonymous
2x dx = dy , so dx = dy/(2x) , so dx = dy/ ( 2 sqrt y)
anonymous
  • anonymous
but it comes out the same
anonymous
  • anonymous
since x^3 = sqrt y ^3
anonymous
  • anonymous
etc
anonymous
  • anonymous
are u a student in college or school?
anonymous
  • anonymous
college
anonymous
  • anonymous
yes thats right..
anonymous
  • anonymous
very nice, both ways work
anonymous
  • anonymous
ok bye.......
anonymous
  • anonymous
bye
anonymous
  • anonymous
i like your way though
anonymous
  • anonymous
it never occured to me you can multiply top and bottom by x

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