Homework Help: Calculus (Partial Fractions)

Posted by Sean on Sunday, February 24, 2008 at 3:25pm.

My Calculus class just started Partial Fractions, and I understand MOST of it.

I'm having a slight problem though, when either C, or C and D come in.

I understand

A/(something) + B/(something)

But I become confused when it's either

A/(something) + B/(something) + C/(something)

OR

A/(something) + B/(something) + (Cx + D)/(something)

Why is it that you get ONLY C, or Cx + D ?

I always get confused when it comes to that, so help would be greatly appreciated.

Calculus (Partial Fractions) - Damon, Sunday, February 24, 2008 at 4:23pm
When you have a quadratic that does not have real roots in the denominator (complex roots only), you put Cx+D in the numerator.
like if your denominator were (x-1)(x+1)(x-2)
you might use
A/(x-1) + B/(x+1) +C/(x-2)
But if your denominator were:
(x-1)(x^2-2x+2)
You could deal with the (x-1) part just fine.
but what to do with the other part?
You can not write it as the sum of two fractions with numerators B and C because you can nor write
x^2-2x+2 as (x+p)(x+q)
so you have to resort to
(C + D x)/(x^2-2x+2)
Calculus (Partial Fractions) - Reiny, Sunday, February 24, 2008 at 4:42pm
This method has practical applications only if the denominator factors.
Since you don't give an example I will supply one

separate (5x^2+3x+4)/(x^3+x^2-2x)

(I started with known fractions and simplified, so that I would have a question that worked out)

The bottom factors to x(x+2)(x-1)

so let
(5x^2+3x+4)/(x^3+x^2-2x) = A/(x+2) + B/(x-1) + c/x

(5x^2+3x+4)/(x^3+x^2-2x)
= [Ax(x-1) + Bx(x+2) + C(x+2)(x-1)]/x(x+2)(x-1)

clearing the denominator we get:
5x^2+3x+4 = Ax(x-1) + Bx(x+2) + C(x+2)(x-1)

now let x=0, then -2C = 4, and C = -2
let x=1, then 3B=12, and B = 4
let x=-2, then 6A = 18 and A = 3

so my original fraction (5x^2+3x+4)/(x^3+x^2-2x) can
be split into
3/(x+2) + 4/(x-1) - 2/x
Calculus (Partial Fractions) - Sean, Monday, February 25, 2008 at 2:28pm
Reiny. You misunderstood. And I didn't word it right.

I meant where does (Cx + D)/Something

Pop up?
Calculus (Partial Fractions) - Anonymous, Monday, February 25, 2008 at 3:11pm
Okay. Example.

If the Denominator is

x^4 - 2x^2 - 8 > (x-2) (x+2) (x^2+2)

Would I do A + B + C, or A + B + (Cx + D)?
Calculus (Partial Fractions) - Sean, Monday, February 25, 2008 at 3:15pm
Wow. I'm making this topic last a while.

I think- and this is just out loud- that you would do (Cx + D) Over the X^2 + 2?

Am I getting that right?

OR

You have Cx + D over something if the SOMETHING has a term of X to a power greater than one?

Answer this Question

Related Questions

Calculus - Integration - I came across this problem in my homework, and I was ...
Calculus - Partial Fractions - What is the integral of 7e^(7t) Divided By e^14t+...
calculus - I am working on this problem and having some trouble. We're ...
Calc II - Perform long division on the integrand, write the proper fraction as a...
Calc easy - Having trouble getting the correct solution. The integral of “x...
Binomial - Help me on this one :( Express y= (7-3x-x^2)/[((1-x)^2)(2+x)] in ...
calculus - what is the integral of (x^2-x+6)/(x^3-3x)? the process involves ...
calculus - what is the integral of x^2-x+6/x^3-3x? the process involves partial ...
calculus - Find the partial fractions for (3x^2-7x+12)/(x-2)(x^2-2x+5)
calculus - Decompose 58-x/x^2-6x-16 into partial fractions.

For Further Reading

First Name:
School Subject:
Answer: