Finding the output of a derivative using integration by partial fractions

In summary, the quadratic function, f(x), with f(0) = 1 and the integral \int \frac{f(x)}{x^2(x+1)^3}dx being a rational function, the value of f '(0) is b = 3.
  • #1
octowilli
10
0

Homework Statement



If f is a quadratic function such that f(0) = 1 and
[tex] \int \frac{f(x)}{x^2(x+1)^3}dx [/tex]
is a rational function, find the value of f '(0).



Homework Equations





The Attempt at a Solution



This question is presented in the context of learning about integration by partial fractions.

For the quadratic bit I have

[tex] f(x) = ax^2 + bx + c [/tex] [tex] f(0) = a(0)^2 + b(0) + c = c = 1 [/tex][tex] f'(x) = 2ax + b [/tex] [tex] f'(0) = 2a(0) + b = b [/tex]

So, I need to find the value of b? Plugging the integral into Wolfram|Alpha gives me

[tex] \int \frac{ax^2+bx+c}{x^2(x+1)^3}dx = -\frac{a-b+c}{2(x+1)^2} + \frac{b-2c}{x+1} + (b-3c)ln(x) - (b-3c)ln(x+1) - \frac{c}{x} + K [/tex]

I'll write out the work for the integral by hand when I submit the assignment. For now, I'm not sure how to relate f '(0) = b with any of this to find b, if it is b that I need to find. Thanks for commenting!
 
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  • #2
If the integral is going to be a rational function, then it shouldn't have any logs in it. What condition will make that true?
 
  • #3
Dick said:
If the integral is going to be a rational function, then it shouldn't have any logs in it. What condition will make that true?

I don't know.

I messed around with this a bit more and here's what I have.

[tex] \frac{ax^2+bx+c}{x^2(x+1)^3} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{(x+1)} + \frac{D}{(x+1)^2} + \frac{E}{(x+1)^3} [/tex]

[tex] ax^2 + bx + c = A(x)(x+1)^3 + B(x+1)^3 + C(x^2)(x+1)^2 + D(x^2)(x+1) + E(x^2) [/tex]

From there, using the method of partial fractions, you would expand, group, and factor, right? Then setup a system of equations?

And here's what I have just integrating with the unknown coefficients.

[tex] \int \frac{ax^2+bx+c}{x^2(x+1)^3} = Aln|x| - \frac{B}{x} + Cln|x+1| - \frac{D}{x+1} - \frac{E}{2(x+1)^2} [/tex]
 
  • #4
Another thought. When x = 0 in the quadratic,

[tex] a(0)^2 + b(0) + c = A(0)(0+1)^3 + B(0+1)^3 + C(0^2)(0+1)^2 + D(0^2)(0+1) + E(0^2) [/tex]

simplifies to

[tex] c = B = 1 [/tex]

I'm not sure what that tells me. I'll think about it.
 
  • #5
Dick said:
If the integral is going to be a rational function, then it shouldn't have any logs in it. What condition will make that true?

Alright I thought about this again. For the integral to be a rational function, the coefficients on the logs need to equal zero.

so

[tex] \int \frac{ax^2+bx+c}{x^2(x+1)^3} = A\cdot ln|x| - \frac{B}{x} + C\cdot ln|x+1| - \frac{D}{x+1} - \frac{E}{2(x+1)^2} [/tex]

simplifies to, given B = 1 above,

[tex] = -\frac{1}{x} - \frac{D}{x+1} - \frac{E}{2(x+1)^2}[/tex].

But I don't think that bit in particular gets me anywhere. However, I can use the fact that A = 0, B = 1, and C = 0 to rewrite

[tex] ax^2 + bx + c = A(x)(x+1)^3 + B(x+1)^3 + C(x^2)(x+1)^2 + D(x^2)(x+1) + E(x^2) [/tex]

as

[tex] ax^2 + bx + c = (x+1)^3 + D(x^2)(x+1) + Ex^2 [/tex].

Now I can expand, group, and factor the right-hand side of the equation to get the following (if I did my work correctly)

[tex] 0x^3 + ax^2 + bx + c = Dx^3 + x^3 + Dx^2 + Ex^2 + 3x^2 + 3x + 1 [/tex] [tex] 0x^3 + ax^2 + bx + c = (D + 1)x^3 + (D + E + 3)x^2 + 3x + 1 [/tex]At this point can I say that b = 3, which is the answer to the problem statement?
 
  • #6
octowilli said:
Alright I thought about this again. For the integral to be a rational function, the coefficients on the logs need to equal zero.

so

[tex] \int \frac{ax^2+bx+c}{x^2(x+1)^3} = A\cdot ln|x| - \frac{B}{x} + C\cdot ln|x+1| - \frac{D}{x+1} - \frac{E}{2(x+1)^2} [/tex]

simplifies to, given B = 1 above,

[tex] = -\frac{1}{x} - \frac{D}{x+1} - \frac{E}{2(x+1)^2}[/tex].

But I don't think that bit in particular gets me anywhere. However, I can use the fact that A = 0, B = 1, and C = 0 to rewrite

[tex] ax^2 + bx + c = A(x)(x+1)^3 + B(x+1)^3 + C(x^2)(x+1)^2 + D(x^2)(x+1) + E(x^2) [/tex]

as

[tex] ax^2 + bx + c = (x+1)^3 + D(x^2)(x+1) + Ex^2 [/tex].

Now I can expand, group, and factor the right-hand side of the equation to get the following (if I did my work correctly)

[tex] 0x^3 + ax^2 + bx + c = Dx^3 + x^3 + Dx^2 + Ex^2 + 3x^2 + 3x + 1 [/tex] [tex] 0x^3 + ax^2 + bx + c = (D + 1)x^3 + (D + E + 3)x^2 + 3x + 1 [/tex]At this point can I say that b = 3, which is the answer to the problem statement?

Yes, f'(0)=b=3. But you didn't need to go through all of that. Look at the integral in your original post. If the log parts are going to vanish you need to have (b-3c)=0. Since you already figured out that c=1. That tells you b=3.
 
  • #7
Dick said:
Yes, f'(0)=b=3. But you didn't need to go through all of that. Look at the integral in your original post. If the log parts are going to vanish you need to have (b-3c)=0. Since you already figured out that c=1. That tells you b=3.

Haha, well, it was fun to do it the long way too! Thanks for helping Dick!
 

What is integration by partial fractions?

Integration by partial fractions is a method used to simplify and integrate rational functions. It involves breaking down a fraction into smaller, simpler fractions that can be easily integrated.

Why is integration by partial fractions useful?

This method is useful because it allows us to integrate rational functions that would be difficult or impossible to integrate using other methods. It also helps us to find the output of a derivative more easily.

What are the steps for finding the output of a derivative using integration by partial fractions?

The steps for finding the output of a derivative using integration by partial fractions are as follows:

1. Factor the denominator of the rational function into linear or irreducible quadratic factors.

2. Write the rational function as a sum of simpler fractions using the partial fraction decomposition method.

3. Integrate each of the simpler fractions separately.

4. Combine the individual integrals to get the final output of the derivative.

What are some common mistakes to avoid when using integration by partial fractions?

Some common mistakes to avoid when using integration by partial fractions include:

- Failing to factor the denominator completely.

- Making errors in the partial fraction decomposition process.

- Forgetting to include the constant of integration when integrating each term separately.

Can integration by partial fractions be used for all rational functions?

No, integration by partial fractions can only be used for proper rational functions, which are those where the degree of the numerator is less than the degree of the denominator. Improper rational functions, where the degree of the numerator is equal to or greater than the degree of the denominator, cannot be integrated using this method.

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