- 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)

- schrodinger

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- anonymous

no

- anonymous

either way worked out

- anonymous

i checked my program

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## More answers

- 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

now go ahead.

- anonymous

yes that works, but i want to use partial fractions

- anonymous

1/(x^2(x^2-15)) = A/x + B/x^2 + (cx+d)/(x^2-15)

- anonymous

thats some trick you got there :)

- anonymous

partial fraction will be avoided when you have better choices...

- anonymous

the tricks will help u and its very time saving...

- anonymous

i guess was wondering, is it ok to have , a quadratic that is not reducible , ok let me state this correctly

- anonymous

how did you know to seperate them like that

- anonymous

only practice..........

- anonymous

i can make it harder, one sec

- anonymous

int (1/ (x^3 (x^2-15))

- anonymous

ok thats a good partial fractions problem

- anonymous

yea...

- anonymous

here you can use partial fractions because no other way............

- anonymous

ok, now we have two options it would seem, my question really boils down to this quadratic

- anonymous

the quadratic is not irreducible,

- 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

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

my question was, both approaches are legal?

- anonymous

does that work, im not sure

- anonymous

both are correct.

- anonymous

oh ok, i thought the quadratic MUST be irreducible

- anonymous

but when u try to find cx+d/(x^2-15) then may be again u have to do partial fractions...

- anonymous

are you sure this is correct 1/15[âˆ«dx/(x(x2âˆ’15)âˆ’âˆ«dx/x3],
i did it and got a different answer

- anonymous

what did u get?

- anonymous

[
x^3 - x(x^2-15) ] / (x^4 (x^2 - 15)

- anonymous

no...

- anonymous

im trying to reverse your result, checking if its equivlanet

- anonymous

((x^2)-(x^2-15))/(x^3(x^2-15))

- anonymous

oh it works

- anonymous

ok

- anonymous

my math was off,

- anonymous

again there is another way to do it just substitute x^2=y.

- anonymous

thats not going to work, since 2x dx = dy

- anonymous

it is again more easier ....

- anonymous

then dx = dy / (2x )

- anonymous

oh you mean for the first problem ?

- anonymous

your integration is xdx/(x^4(x^2-15)

- anonymous

for the second one...

- anonymous

try it... it surely works..

- anonymous

oh , but its not simple

- anonymous

lets try it for the first integral

- anonymous

oh ok , so s

- anonymous

its not for the first one.... its only applicable for the second one....

- anonymous

these some shortcut to approach....

- anonymous

i have 2 go now.....

- anonymous

dx = dy / ( 2 sqrt y )

- anonymous

do you agree to that?

- anonymous

have u understood properly?

- anonymous

yes, i just want to make sure i understand your subsitution

- anonymous

dx/(x^3(x^2-15) gives you dy/(2y^2(y-15))

- anonymous

i get dy / ( 2 y^(3/2) (y-15)

- anonymous

y=x^2
dy=2xdx
dx/(x^3(x^2-15))= 2xdx/(2x^4(x^2-15)) = dy/(2y^2(y-15))

- anonymous

got it?

- anonymous

thats wrong

- anonymous

you have to solve for dx

- anonymous

what?

- anonymous

check it , thats correct...

- anonymous

2x dx = dy , so dx = dy/(2x)

- anonymous

ok...let me to make u clear...

- anonymous

oh, you multiplied top and bottom by 2x, one sec

- anonymous

yesssssssssss

- anonymous

be fast i have to go .... but its nice to solve maths with you?

- anonymous

very cleverl

- anonymous

thanks , bye

- anonymous

sure we will talk again

- anonymous

thats the magic of maths...........

- anonymous

my way probably works, i have to work on it :)

- anonymous

my substitution

- anonymous

what do u do?

- anonymous

i solve for dx =

- anonymous

from where r u?

- anonymous

i live in usa, near new york

- anonymous

in which class are u?

- anonymous

2x dx = dy , so dx = dy/(2x) , so dx = dy/ ( 2 sqrt y)

- anonymous

but it comes out the same

- anonymous

since x^3 = sqrt y ^3

- anonymous

etc

- anonymous

are u a student in college or school?

- anonymous

college

- anonymous

yes thats right..

- anonymous

very nice, both ways work

- anonymous

ok bye.......

- anonymous

bye

- anonymous

i like your way though

- anonymous

it never occured to me you can multiply top and bottom by x

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