can someone help me with partial fraction ? 4/x(x^2+4)

i might

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

yes thats right

yes thats right

cool, so what's the first step? separate the function based upon the denominator, you have two functions mutliplied by eachother on the bottom: x * (x^2+4)

so you could create a sum of two fractions: one with x as the denonm. and one with (x^2+4) as the denom. can you post those please?

i don't quite get it.. ><

np, ok, the goal is to express the function as the sum of other functions, which are actually "fractions"...where the denominators of these functions that you want to sum are the parts of the denominator of the larger function....clear?

so basically, here you have 4 ?( x * (x^2 + 4) )...the larger function on the denom is (x * (x^2 + 4) )....so can you separate this into two functions?

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

right now, ignore the numerator so you're right, but ignore 4...the setup is now: \[4/(x * (x^2 + 4) ) = A/(x) + B/(x^2+4)\]

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)?

sorry, you are right about that equation not being quite right, because there is x^2+4...so you actually need to adjust the numerator of the fraction with (x^2+4)

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

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

ok, now do you think you can make these fractions so that they have a common denominator? hint: multiply by "1"

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

ok, however you are adding these fractions...so you cannot cross multiply...you need to "give" the fractions the same denominator...and easy way would be multiply each fraction by 1, where 1 is actually a fraction with the numerator and denominator equal, in this case you would want to multiply A/x by (x^2+4)/(x^2+4)...see how this multiplying by 1 is OK?

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?

right now you've only multiplied the second fraction by (1/x)...so you didn't multiply by "1" like you did for the first with (x^2+4)/(X^2+4)

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)?

not quite, right it is multiplied by 1/x...see how you put x in the denominator? multiply the numerator by x too...so that you have multiplied by "1"..right?

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

actually do the opposite....so (Bx+c)/(x^2+4) becomes ((Bx+c)*x)/((x^2+4)*x)...see how you multiplied numerator and denominator by x because you mutliplied the fraction by "1" where "1" = x/x

so, what are the two fractions now, now that they have common denominators of x*(x^2+4)...rewrite the equation

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

right on

so the entire equation looks like this: \[4/(x * (x^2 + 4)) = ((A * (x^2 + 4)) + ((Bx + c) *(x)))/(x * (x^2 + 4))\]

yeap..

nice, now you can ignore denominators for the time being...so we actually want to look at the equality between the numerators

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

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.

cross multiply refers to when you have two fractions: like a/b = c/d , cross multiplying would yield a*d = c*b...

that is what cross-multiplying technically is...it is best to the procedure as above so as to not move too hastily and miss a step...

ohh..

so anyway, we are close to solving

4=(A∗(x2+4))+(Bx+c)∗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, now, let's consider this, both equations are assumed to be equal for all values of x...right, because it's an equation

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?

but as from just now eqn 4 = A(x^2+4) + x(Bx+c) can i sub X = 0 for Bx+c(x) so that 4=A(0^2+4)? which turn out to be 4 = 4A?

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

now, on the right side here there are no terms with x^2 nor x...so what do you think C would be equal to?

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..

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