Partial Fraction Problem

[Jason76]

Prove A = 0, B = 3, C = 4

\(\displaystyle \int \dfrac{x^{3} + 6x^{2} + 3x + 6}{x^{3} + 2x^{2}} dx\)

This becomes \(\displaystyle 1 + \dfrac{4x^{2} + 3x + 6}{x^{3} + 2x^{2}}\) How did this line come from the one above?

\(\displaystyle \dfrac{4x^{2} + 3x + 7}{x^{3} + 2x^{2}}\)

\(\displaystyle \dfrac{4x^{2} + 3x + 7}{x^{3} + 2x^{2}} = \dfrac{A}{x} + \dfrac{B}{x^{2}} + \dfrac{C}{x + 2}\)

\(\displaystyle \dfrac{4x^{2} + 3x + 7}{x^{2}(x + 2)} = \dfrac{A}{x} + \dfrac{B}{x^{2}} + \dfrac{C}{x + 2}\)

\(\displaystyle 4x^{2} + 3x + 7 = Ax(x + 2) + B(x + 2) + Cx^{2}\)

\(\displaystyle 4(0)^{2} + 3(0) + 7 = A(0)((0) + 2) + B((0) + 2) + C(0)^{2}\)

\(\displaystyle 4(-2)^{2} + 3(-2) + 7 = A(-2)((-2) + 2) + B((-2) + 2) + C(-2)^{2}\)

 

[Jason76]

I think a better way to do this is:

Prove A = 0, B = 3, C = 4

\(\displaystyle \int \dfrac{x^{3} + 6x^{2} + 3x + 6}{x^{3} + 2x^{2}} dx\)

This becomes \(\displaystyle \dfrac{x^{3} + 6x^{2} + 3x + 6}{x^{3} + 2x^{2}}\)

\(\displaystyle \dfrac{x^{3} + 6x^{2} + 3x + 6}{x^{3} + 2x^{2}}\)

\(\displaystyle \dfrac{x^{3} + 6x^{2} + 3x + 6}{x^{3} + 2x^{2}} = \dfrac{A}{x} + \dfrac{B}{x^{2}} + \dfrac{C}{x + 2} = 4\)

\(\displaystyle \dfrac{x^{3} + 6x^{2} + 3x + 6}{x^{2}(x + 2)} = \dfrac{A}{x} + \dfrac{B}{x^{2}} + \dfrac{C}{x + 2}\)

\(\displaystyle x^{3} + 6x^{2} + 3x + 6 = Ax(x + 2) + B(x + 2) + Cx^{2}\)

\(\displaystyle (0)^{3} + 6(0)^{2} + 3(0) + 6 = A(0)((0) + 2) + B((0) + 2) + C(0)^{2} = 4\)

\(\displaystyle (-2)^{3} + 6(-2)^{2} + 3(-2) + 6 = A(-2)((-2) + 2) + B((-2) + 2) + C(-2)^{2}\)

 

[Deleted member 4993]

Prove A = 0, B = 3, C = 4

\(\displaystyle \int \dfrac{x^{3} + 6x^{2} + 3x + 6}{x^{3} + 2x^{2}} dx\)

This becomes \(\displaystyle 1 + \dfrac{4x^{2} + 3x + 6}{x^{3} + 2x^{2}}\) How did this line come from the one above?

Using division of polynomials

\(\displaystyle \dfrac{4x^{2} + 3x + 6}{x^{3} + 2x^{2}}\)

\(\displaystyle \dfrac{4x^{2} + 3x + 6}{x^{3} + 2x^{2}} = \dfrac{A}{x} + \dfrac{B}{x^{2}} + \dfrac{C}{x + 2}\)

An easy way to find C

\(\displaystyle \dfrac{4x^{2} + 3x + 6}{x^{3} + 2x^{2}} = \ \dfrac{Ax + B}{x^{2}} + \dfrac{C}{x + 2}\)

Multiply both sides by x + 2
\(\displaystyle \dfrac{4x^{2} + 3x + 6}{x^2} = \ \dfrac{(Ax + B)(x+2)}{x^{2}} + C\)

Find limit of both sides as x → (-2)



\(\displaystyle \dfrac{16 - 6 + 6}{4} = C \)


C = 4


Now find A and B

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[Jason76]

So is long division necessary in this case, or is it just an option, and why?

 

[stapel]

So is long division necessary in this case, or is it just an option, and why?

Dunno. Are you going to answer the question implicitly posed in the first reply, in which it was pointed out that you haven't told us what the actual question is?

You've told us your answer (I think) for the values of A, B, and C, and have told us to prove your answer, and seem now to be quizzing us on what the "correct" method of proof is, but you haven't yet told us if we've guessed your meaning correctly. Seriously, we really can't read your mind.

 

[Deleted member 4993]

So is long division necessary (yes) in this case, or is it just an option, and why (look at the form of the "fractions" you have chosen and tell us "why") ?

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