Partial Fraction Decomp/ How do you know when to use C/equation, VS making CX+D the numerator in an equation?

Question

EG: x+5/x^2(x^2+4) = (A/x )+( B/x)+ (Cx+D)/(x^2+4)
Vs x-1/(x+3)(x+5)^2 = (A/x+3)+ B/(x+5) +C/(x+5)^2

I am confused on the final step, in deciding when I should use a single variable (eg B/C/D) as the numerator in the last fraction, vs using Cx+D/ Bx+C (etc).

Can someone please tell me what purpose in life partial fraction domposition serves??? Variables are not solved, and it seems like an action in futility, simply rewriting a variable equation in a different format.

Thank you for your help and patience.

Answer 1

You use the Cx+D numerator for a denominator with a quadratic form, ax^2+bx+c. This is true of your first example. You just use the A, B, C numerator if the denominator of the fraction is of the form (x-a)^n, like (x-1), (x-1)^2, (x-1)^3. Even though (x-1)^2 can be written into a quadratic, is still has the form of (x-a)^n, just like in your 2nd example.



In terms of purposes, Partial Fraction Decomposition is important for several Engineering processes. P.F.D. is essential for utilizing Laplace and Z-transforms, which are essential tools for analog and digital signal analysis and processing.



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Hope this helps

Answer 2

Partial fraction decomposition makes it possible to solve integrals of a certain type.



For instance, there's no way I can solve the second integral you gave in its current form, but I CAN solve it using the second form.



Integral (A/(x+3) + B/(x+5) + C/(x+5)^2)dx



We can separate this into three integrals



Integral (A/(x+3))dx + Integral (B/(x+5))dx + Integral (C/(x+5)^2)dx



And then pull out the constants (since they get in our way of integration sometimes).



A * Integral (1/(x+3))dx + B * Integral (1/(x+5))dx + C * Integral(1/(x+5)^2)dx



I *know* the integral of 1/(x+3); that's just ln |x+3|. Also, I know the integral of 1/(x+5)^(2); since x + 5 is linear, we just apply the reverse power rule for derivatives, and convert this to (x+5)^(-2), to get (x+5)^(-1)/(-1), or -1/(x+5). So now our answer is



A * ln|x+3| + B ln |x+5| + C (-1/(x+5)) + D for some constant D.



Notice that using partial fraction decomposition makes integration actually possible. There's no way we can solve the original question by u substitution, parts, or any of the other existing rules. We have to decompose it into smaller parts which we can actually solve.



The second step is to solve for A, B, and C. We know that



(x-1)/(x+3)(x+5)^2 = (A/x+3)+ B/(x+5) +C/(x+5)^2



is true for _all_ x. But first, let's multiply both sides by (x+3)(x+5)^2. This cancels out all the denominators, and we're left with:



x - 1 = A(x+5)^2 + B(x+3)(x+5) + C(x+3)



Since this is true for all x, let's let x = -5. Then



-5 -1 = A(0)^2 + B(-2)(0) + C(-2)

-6 = C(-2)

C = 1/3



So we at least have our C value now.

Now, let x = -3. Then



-3 - 1 = A(-3+5)^2 + B(0)(2) + C(0)

-4 = A(2)^2

-4 = 4A

A = -1



To solve for B is a bit trickier. We have to actually multiply everything out.



x - 1 = A(x+5)^2 + B(x+3)(x+5) + C(x+3)

x - 1 = A(x^2 + 10x + 25) + B(x^2 + 8x + 15) + Cx + 3C

x - 1 = Ax^2 + 10Ax + 25A + Bx^2 + 8Bx + 15B + Cx + 3C



And now, we group together like terms



x - 1 = (A+B)x^2 + (10A + 8B + C)x + (25A + 15B + 3C)



Now, we compare it to the left hand side, and equate it. By the rules of partial fractions, the coefficients on the left hand side and the right hand side MUST match. There's no x^2, so we assume there's a 0x^2 on the left hand side.



0x^2 + x - 1 = (A+B)x^2 + (10A + 8B + C)x + (25A + 15B + 3C)



And now we match:



A+B = 0

10A + 8B + C = 1

25A + 15B + 3C = -1



Since we solved for A earlier (it was -1), we can solve for the unknown B, using the first and simplest equation.



-1 + B = 0, so B = 1



So now, A = -1, B = 1, and C = 1/3



Now, we plug them back into here:



A * ln|x+3| + B ln |x+5| + C (-1/(x+5)) + D



(-1)ln|x+3| + (1) ln|x+5| + (1/3) (-1/(x+5)) + D

- ln |x+3| + ln|x+5| - (1/3) (1/(x+5)) + D





To answer your other question, when to use Cx + D in the numerator, it is when the partial fraction decomposition creates something that's not factorable.



For instance, if you have x^2 + 4, you can't factor it, and so you use (Cx + D)/(x^2+4). Had it been x^2 - 4, then this is something you CAN factor, as it becomes (x-2)(x+2) and then you do your A/(x-2) + B/(x+2) as normal.

Answer 3

I had a similar problem..and have written the answers in my notes somewhere but cant find them...

they are basically rules u have to follow like when you have (x+2) ^2 in the denominator you will have B/(x+5) +C/(x+5)^2..the best bet is that if you dont get an answer on the forum, refer to worked out examples and you should be able to figure out the rules through intuition..or check out a text book..



if you go ahead with a uni degree which is to do with a numerate discipline..you will come across partial fractions again and again..its used extensively in things like laplace transforms...or basically whenever you require to break down a function to study its behaviour...by using partial fractions you effectively break down a function into its smaller components..and each component can be studied to find out the total effect of the function..

Answer 4

(x+5)/[(x^2)(x^2+4)]



breaks down to...



(x+5)/(x^2)+ (x+5)/(x^2+4)



When you have x in the denominator, you use just one letter. If you have x^2 (such as this case) you use (A/x)+(B/X), because x^2 is simply x*x, so you can break it up further.



When you have (x^2+4) in the denominator, you use (Ax+B) or (Cx+D) in this example, because you cannot break it up further and its a quadratic equation.



Simple way to remember:



If the denominator is broken up as much as you can and is still a quadratic, use (ax+b).



x=A

x^2+1= Ax+b

x^3+1=Ax^2+Bx+C



You get the pattern.