Question

Hello(: So I'm trying to use partial fractions on this prob. I factored the denominator to x and (x^2+1).
Then i put A/x + (Bx+C)/(x^2+1)....is this incorrect? Cuz it doesn't seem to be working<

**postin prob

Answer 1

anyone get this? I have a test in a couple hours<< i was absent for this lesson so I don't really now how to go about for the more "seemingly" complicated probs

Answer 2

@ganeshie8 do u think u can take a shot at this if u get a minute?

Answer 3

thats right !

Answer 4

kk and now I'm trying to use i to solve for B and C.

Answer 5

did u solve A already ?

Answer 6

yes A would be 4 right?

Answer 7

yup ! did u use coverup method ?

Answer 8

what's the coverup method? I never heard of it. Could be cuz i wasn't in class so I haven't been exposed to to the technical terms lol

Answer 9

no need of using coverup method, forget it :)
do it in ur own way, all u want is to find A, B, and C values. thats all

Answer 10

3x^2 + 4x +4 = A(x^2+1) + (Bx+C)x

Answer 11

compare x^2, x and constant terms on both sides,
u will get 3 equations, u can slve them

Answer 12

haha okay got that. how would I get 3 equ? I got 2 but can't workout the 3rd. For my first i plugged in x=0 (to get A) and then I plugged in x=i. what would i do for the last one?

Answer 13

the standard method is to compare x^2, x and constant terms both sides

Answer 14

could u elaborate a bit further if u don't mind? lol sorry its 4 in the morning, haha my brain isn't with me

Answer 15

3x^2 + 4x +4 = A(x^2+1) + (Bx+C)x

x^2 coefficcient on left side = 3
x^2 coefficient on right side = A + B

so, 3 = A + B ----------- (1)

Answer 16

thats ur first equation

Answer 17

comparing x coefficient both sides :
4 = C --------------(2)

Answer 18

comparing constant term both sides :
4 = A ------------(3)

Answer 19

you have 3 equaitons wid 3 unknowns
u can solve them :)

Answer 20

ok! Ive never compared them to get equations like that. Thanks for the hints!!

Answer 21

this method always works.

i prefer this, to plugging a value randomly..

Answer 22

lets do one more problem, im sure u will feel this method is easy :)

Answer 23

OMG LOL just got it! Wow thank you!! So much more easier lol! Appreciate it!!!!

Answer 24

np :) good luck wid the exam !

Answer 25

just one important thing i wana tell, since u missed the class for partial fractions

Answer 26

thank you! it's like magic haha

Answer 27

when integrating partial fractions work ONLY if below condition satisfies :
degree of numerator < degree of denominator

Answer 28

if u have the integral like below :


x^3 + x
------------
x^2 + 2x + 1

you cannot use partial fractions directly; u need to do long division first, and make the degree of numerator LESS than degree of denominator

Answer 29

and u said it always works right? like there is no conditions for that little trick u used?

and i got that. If the num is greater u divided it out then integrate what u get

Answer 30

yes, the 'comparing coefficients' method works always.

Answer 31

u deserve some good sleep before exam lol... you should sleep a bit :)

Answer 32

perfect! Thanks again!! Really appreciate ur time!

Answer 33

np :)

Answer 34

Have a nice day(: And I just had to figure out how to do that prob before i could sleep.

Answer 35

you will get ln's, and inverse tans whenever u do partial fractions.. have fun :)