Derivatives of partial fractions

In summary: Bx(x-3) + C(x-3)In summary, the speaker is having trouble solving an integral problem that involves breaking it down into partial fractions. They have found the equations for A, B, and C, but are unsure if they are correct. They have even checked their answers with Wolfram|Alpha, but are still being marked as incorrect.
  • #1
Badgerspin
15
0
I'm having issue with one problem. We're asked to break down the problem into partial fractions to solve for the integral.

Well, I'm stuck on one. I'm being asked for the values of A, B, and C for the following problem.

∫((9x^2+13x-83)/((x-3)(x^2 + 16)))dx

I can get it worked down pretty far, but I'm continually given back that I've done it incorrectly.

Broken down:

(9x^2+13x-83)/((x-3)(x^2 + 16)) = A/(x-3) + (bx + c)/(x^2 + 16)

From there, I get (multiplying both sides by the common denominator, and now only working with the right side as it's the only one that needs to be worked):

A(x^2+12) + Bx(x-3) + C(x-3)
= ax^2 + 12A + Bx^2 - 3Bx + Cx - 3C
= (A+B)x^2 + (C-3B)x + (12A - 3C)

Therefor:

-9x^2 + 13x -83 = (A+B)x^2 + (C-3B)x + (12A - 3C)

Breaking it down further, I get:

A + B = -9
3B - C = 13
12A - 3C = -83

I'm thinking something may be wrong with my values in the second part. Any help would be appreciated. Whatever answers I put in, I'm marked as incorrect. I went so far as to plug the last little part I posted into Wolfram|Alpha to confirm, and it spits the same answers I've been getting. I get the feeling that I did something incorrectly earlier on.
 
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  • #2
Badgerspin said:
I'm having issue with one problem. We're asked to break down the problem into partial fractions to solve for the integral.

Well, I'm stuck on one. I'm being asked for the values of A, B, and C for the following problem.

∫((9x^2+13x-83)/((x-3)(x^2 + 16)))dx

I can get it worked down pretty far, but I'm continually given back that I've done it incorrectly.

Broken down:

(9x^2+13x-83)/((x-3)(x^2 + 16)) = A/(x-3) + (bx + c)/(x^2 + 16)

From there, I get (multiplying both sides by the common denominator, and now only working with the right side as it's the only one that needs to be worked):

A(x^2+12) + Bx(x-3) + C(x-3)
I think the mistake is here, it should be A(x^2+16)
 

Related to Derivatives of partial fractions

1. What are derivatives of partial fractions?

Derivatives of partial fractions are a mathematical technique used to find the rate of change of a function that is expressed as the sum of simpler fractions.

2. How are derivatives of partial fractions calculated?

To calculate the derivative of a partial fraction, each individual fraction is differentiated separately using the power rule, product rule, or chain rule, depending on the form of the fraction.

3. What is the purpose of using derivatives of partial fractions?

The purpose of using derivatives of partial fractions is to simplify and solve complex integration problems by breaking down a function into smaller, more manageable fractions.

4. Can derivatives of partial fractions be used for any type of function?

No, derivatives of partial fractions are specifically used for integrals involving rational functions, which are functions that can be expressed as a ratio of two polynomials.

5. Are there any limitations to using derivatives of partial fractions?

One limitation is that derivatives of partial fractions can only be used for functions that can be expressed as the sum of simpler fractions. Additionally, this method can be time-consuming and tedious for more complex functions.

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