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i might

so the function is 4/( x * ( x^2 + 4) )?

yes thats right

yes thats right

i don't quite get it.. ><

4 / x + 1/(x^2+4) im not very sure.. i suck at maths

does that look familiar, like something in your text?

but shouldn't it be A / x + Bx+c / (x^2+4)?

ok, why do you think that you use "x" twice?

because of the demo has a (x^2)?

good catch! the equation becomes:
\[4/(xâˆ—(x2+4))=A/(x)+(Bx+C)/(x2+4) \]

yeap.. that about it.. the only thing is i don't know how to solve for Bx+C

i did a cross multipy which end up to be 4= A(x^2+4) + X(Bx+c)?

so the fraction on becomes: \[( A * (x^2 + 4))/(x * (x^2 +4))\]

cool?

ohh icic

nice, now can you apply the concept to the other fraction? then post the new equation?

A(x^2+4) / x(x^2+4) + Bx+C / x(x^2+4) am i right?

almost, what about the numerator on the second fraction?

the second fraction demo is x(x^2+4) isnt it same as the 1st fraction 's demo?

yes, that's right, but the numerator is missing something...

\[(A * (x^2 + 4) ) /( x * x^2 +4 )) + ((Bx + C)*(????))/(x * (x^2 + 4))\]

is it 1 ?

why 1?

then is it Bx+C * x(x^2+4)?

so you're close, right?

so i got to muplity Bx+C with X(x^2+4)?

Bx+c * 1/x / x(x^2+4) ?? im confused

no, you were very close with (Bx + c) /x * (x^2 + 4)...how did you get here from (Bx+c)/(x^2 + 4)?

oh cancel out the x from the x(9x^2+4)
but then the 1st fraction's demo isnt the same as 2nd one?

typo error x(x^2+4)

er, no canceling now

A( x^2+4) / x(x^2+4) + Bx+c * (x) / x (x^2+4)

right on

yeap..

\[4 = (A *(x^2 + 4)) + (Bx+c)*x \]

this eqn isnt it same as what just now i had cross multiply them together?

i'm not sure what you mean by cross multiply...

as stated just now A/x + Bx+c/ x^2+4
i cross mutiply them.. meaning. A * (x^2+4) and x* (bx+c)

in the original equation you're saying that you multiplied everything by the denominator?

yea.

ohh..

so anyway, we are close to solving

4=(Aâˆ—(x2+4))+(Bx+c)âˆ—x

\[4 = (A * (x^2 + 4)) +(Bx + c)*(x)\] expand this

4 = Ax^2 + 4A + Bx^2 + cx?

\[4 = (A * (x^2 + 4)) + (Bx + C) * x = Ax^2 + 4A + Bx^2 + Cx\]

so you see how only one term does not have x in it?

the 4A

right on! so now we need to figure out what each letter should equal

therefore A = 1

dude, nice!

yeah, the simplest case was to consider that x = 0, now we have that A =1...so let's find C

there are no x terms on the left side, so, what do you think C equals?

Bx^2+Cx...that is the idea

Ax^2+4A+Bx^2+Cx, ok we should group these with like terms -->
(A+B)*x^2 +(C)*x + 4A

(A+B)*x^2 +(C)*x + 4A = 4

you figured out that A = 1 by looking at the case when x=0

yea

and that where i struck finding b and c

C = 0?

right on!

now, you know A=1, C=0, now...to find B....so we look at where we have B, only ing (A+B)*x^2

which (1 + B) *x^2 because A=1

again, there are no x^2 terms on the other side..so B must equal what?

3?

why 3?

the idea is that on the other side the term is 4 + 0*x^2, so we have (1+B)*x^2 = (0)*x^2

ohh.. its -1

yes, so what are A, B and C

A= 1
B= -1
C= 0

one you have them all, substitute them back in the fractions

oicic

where they were separted

in here: 4/(x*(x^2+4))=A/(x)+B/(x^2+4)

ok got it.. thx a lot man~
sry to wate such a long time.. hope u don't mind

no problem, now do you think you can solve another partial fraction problem?

not really.. cos' theres a lot of type of partial fraction..

just use this conversation as notes....try another and good luck!

ok thanks a lot

np