Question

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)

Answer 1

no

Answer 2

either way worked out

Answer 3

i checked my program

Answer 4

\[1/15\int\limits_{}^{}(x ^{2}-(x ^{2}-15))dx/(x ^{2}(x ^{2}-15))\]\[1/15[\int\limits_{}^{}dx/(x ^{2}-15)-\int\limits_{}^{}dx/x ^{2}\]

Answer 5

now go ahead.

Answer 6

yes that works, but i want to use partial fractions

Answer 7

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

Answer 8

thats some trick you got there :)

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

how did you know to seperate them like that

Answer 13

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

Answer 14

i can make it harder, one sec

Answer 15

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

Answer 16

ok thats a good partial fractions problem

Answer 17

yea...

Answer 18

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

Answer 19

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

Answer 20

the quadratic is not irreducible,

Answer 21

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)

Answer 22

here u can use the one which i gave first to reduce it . it would give you \[1/15[\int\limits_{}^{}dx/(x(x ^{2}-15)-\int\limits_{}^{}dx/x ^{3}]\]

Answer 23

my question was, both approaches are legal?

Answer 24

does that work, im not sure

Answer 25

both are correct.

Answer 26

oh ok, i thought the quadratic MUST be irreducible

Answer 27

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

Answer 28

are you sure this is correct 1/15[∫dx/(x(x2−15)−∫dx/x3],
i did it and got a different answer

Answer 29

what did u get?

Answer 30

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

Answer 31

no...

Answer 32

im trying to reverse your result, checking if its equivlanet

Answer 33

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

Answer 34

oh it works

Answer 35

ok

Answer 36

my math was off,

Answer 37

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

Answer 38

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

Answer 39

it is again more easier ....

Answer 40

then dx = dy / (2x )

Answer 41

oh you mean for the first problem ?

Answer 42

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

Answer 43

for the second one...

Answer 44

try it... it surely works..

Answer 45

oh , but its not simple

Answer 46

lets try it for the first integral

Answer 47

oh ok , so s

Answer 48

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

Answer 49

these some shortcut to approach....

Answer 50

i have 2 go now.....

Answer 51

dx = dy / ( 2 sqrt y )

Answer 52

do you agree to that?

Answer 53

have u understood properly?

Answer 54

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

Answer 55

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

Answer 56

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

Answer 57

y=x^2
dy=2xdx

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

Answer 58

got it?

Answer 59

thats wrong

Answer 60

you have to solve for dx

Answer 61

what?

Answer 62

check it , thats correct...

Answer 63

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

Answer 64

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

Answer 65

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

Answer 66

yesssssssssss

Answer 67

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

Answer 68

very cleverl

Answer 69

thanks , bye

Answer 70

sure we will talk again

Answer 71

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

Answer 72

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

Answer 73

my substitution

Answer 74

what do u do?

Answer 75

i solve for dx =

Answer 76

from where r u?

Answer 77

i live in usa, near new york

Answer 78

in which class are u?

Answer 79

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

Answer 80

but it comes out the same

Answer 81

since x^3 = sqrt y ^3

Answer 82

etc

Answer 83

are u a student in college or school?

Answer 84

college

Answer 85

yes thats right..

Answer 86

very nice, both ways work

Answer 87

ok bye.......

Answer 88

bye

Answer 89

i like your way though

Answer 90

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