Integral Help: Solving a Complex Problem

In summary, the integral of (x^3-5x)/(x^3-2x^2+4x-8) is equal to x + (9/8)log| x^2+4 | - (9/4)arctan(x/2) - (1/4)log| x-2 | + C, where C is a constant. This can be found by decomposing the fraction into simpler fractions using the method of partial fractions and then integrating each term separately.
  • #1
tylersmith7690
21
0
Integral help!

Homework Statement


Evaluate the following inde nite integral,
∫[itex]\frac{x^3-5x}{x^3-2x^2+4x-8}[/itex] dx


Homework Equations


I think I am right until i get to equating the equations when I do it the method I am taught in class I get c=0 and I don't think that is right. However when i put it into a system of equations with the coefficients i get fractions.

I would like to know what to do next. Thanks



The Attempt at a Solution



First the degree of numerator is same as the denomenator. So i do long division and end up with

∫1+ [itex]\frac{2x^2-9x+8}{x^3-2x^2+4x-8}[/itex]

Now decompose the fraction part into:

[itex]\frac{2x^2-9x+8}{(x^2+4)(x-2)}[/itex] =[itex]\frac{Ax+B}{(x^2+4)}[/itex]+[itex]\frac{C}{(x-2)}[/itex]

Then get a common denomenator for RHS
= [itex]\frac{(Ax+B)(x-2)+C(x^2+4)}{(x^2+4)(x-2)}[/itex]

From here i tried if x=2 to cancel out the (Ax+B) term which results in C=0
this is where i believe an error has occured.

My other thinking was can I expand then factor out the x^2 and x and constant leaving the A and B and C in whatever they factor into and then do a system of equations.
My result this way is
A= 9/4, B =-9/2, C=-1/4

Kind regards
 
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  • #2
tylersmith7690 said:
From here i tried if x=2 to cancel out the (Ax+B) term which results in C=0
this is where i believe an error has occured.

My other thinking was can I expand then factor out the x^2 and x and constant leaving the A and B and C in whatever they factor into and then do a system of equations.
My result this way is
A= 9/4, B =-9/2, C=-1/4

Those values of A,B and C are correct. However, in your initial attempt, you should not get C=0.
 
  • #3
tylersmith7690 said:

Homework Statement


Evaluate the following inde nite integral,
∫[itex]\frac{x^3-5x}{x^3-2x^2+4x-8}[/itex] dx


Homework Equations


I think I am right until i get to equating the equations when I do it the method I am taught in class I get c=0 and I don't think that is right. However when i put it into a system of equations with the coefficients i get fractions.

I would like to know what to do next. Thanks



The Attempt at a Solution



First the degree of numerator is same as the denomenator. So i do long division and end up with

∫1+ [itex]\frac{2x^2-9x+8}{x^3-2x^2+4x-8}[/itex]

Now decompose the fraction part into:

[itex]\frac{2x^2-9x+8}{(x^2+4)(x-2)}[/itex] =[itex]\frac{Ax+B}{(x^2+4)}[/itex]+[itex]\frac{C}{(x-2)}[/itex]

Then get a common denomenator for RHS
= [itex]\frac{(Ax+B)(x-2)+C(x^2+4)}{(x^2+4)(x-2)}[/itex]
So the two numerators are equal: [itex]2x^2- 9x+ 8= (Ax+B)(x- 2)+ C(x^2+ 4)[/itex]

From here i tried if x=2 to cancel out the (Ax+B) term which results in C=0
this is where i believe an error has occured.
Taking x= 2 gives 8- 18+ 8= 8C which gives C= -1/4, as you have below, not 0. Did you forget the left side of the equation?

My other thinking was can I expand then factor out the x^2 and x and constant leaving the A and B and C in whatever they factor into and then do a system of equations.
My result this way is
A= 9/4, B =-9/2, C=-1/4

Kind regards
If, in [itex]2x^2- 9x+ 8= (Ax+B)(x- 2)+ C(x^2+ 4)[/itex], you take x= 0 (just because it is an easy number) you get [itex]8= (0+ B)(-2)+ C(4)[/itex] or [itex]-2B= -8- (-1/4)(4)= -9[/itex] so that C= -9/2, again as you say. Finally, taking x= 1 (again just because it is an easy number) gives [itex]2- 9+ 8= (A+ B)(-1)+ C(5)[/itex] or [itex]1= -A+ 9/2- 5/4[/itex] so that [itex]A= 9/2- 1- 5/4= 18/4- 4/4- 5/4= 9/4[/itex], again, exactly what you have.

Great! Now all you need to do is the integral:
[tex]\int dx+ \dfrac{9}{4}\int \dfrac{xdx}{x^2+ 4}- \dfrac{9}{2}\int \dfrac{dx}{x^2+ 4}- \dfrac{1}{4}\int \dfrac{dx}{x- 2}[/tex]
 
  • #4
Thanks for the reply, looking again at how i got C=0 was most likely due to it being very late at night :).
But yes I find it easier solving for A,B,C in the system of equations. I see now that my thinking was on the right track and it was just the small error that through me off. Thanks for the help.
 
  • #5
Is my final answer correct?

x + [itex]\frac{9}{8}[/itex] log | x2+4 | - [itex]\frac{9}{4}[/itex] arctan([itex]\frac{x}{2}[/itex]) - [itex]\frac{1}{4}[/itex] log | x-2 | + C
 

1. What is Integral Help?

Integral Help is a problem-solving approach that involves analyzing and addressing complex problems by considering all of the interrelated parts and systems involved. It takes a holistic and integrative approach, rather than focusing on individual elements, to find effective solutions.

2. How does Integral Help differ from other problem-solving methods?

Integral Help differs from other problem-solving methods in that it considers the whole system and all of its parts, rather than just focusing on one aspect. It also takes into account the dynamic and interconnected nature of complex problems, rather than treating them as static and isolated issues.

3. What are the main benefits of using Integral Help?

The main benefits of using Integral Help include a more comprehensive understanding of the problem, the ability to identify and address underlying causes rather than just symptoms, and a more effective and sustainable solution. It also promotes collaboration and communication among stakeholders.

4. Can Integral Help be applied to any type of complex problem?

Yes, Integral Help can be applied to a wide range of complex problems, including social, environmental, and organizational issues. It is a flexible and adaptable approach that can be tailored to fit the specific needs and complexities of each problem.

5. How can I learn more about Integral Help and how to apply it?

There are various resources available for learning about Integral Help, including books, articles, and workshops. You can also consult with professionals who have experience in using this approach. It is important to have a solid understanding of the principles and techniques involved before applying Integral Help to a complex problem.

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