Partial fraction Urgent

  • Thread starter bayan
  • Start date
202
0
Hi every one.

Just came acroos a nasty piece and was wondering if you could help me with it.

I wasn't sure if it belong to math section or here but scince it is for my homework I placed it here.
Here is the question.
Find the integral of [tex]\frac{x}{(x+1)(x-2)^2}[/tex]
I have gotten to some piont using the partial fraction.
here is my work so far.

[tex]\frac{A}{(x+1)}+\frac{B}{(x-2)}+\frac{C}{(x-2)^2}[/tex]

I found the values of A B and C.

C=[tex]\frac{2}{3}[/tex] B=[tex]\frac{2}{3}[/tex] A=[tex]\frac{21}{9}[/tex]

If you could be so kind and help me through with it by typing what you have done would be really nice, I am not asking in latex form a simple typing will do the job.

Also can anyone help me finding the equation of the graph attached?
 

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bayan said:
If you could be so kind and help me through with it by typing what you have done would be really nice, I am not asking in latex form a simple typing will do the job.
No-one is going to do your work for you. You've done the partial fractions, so what's wrong with doing the integration?
 
2,208
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If you've found A B and C ccorrectly (havent checked yet) then you can just substitue your new expression into hte integral.

[tex] \int \frac{x}{(x+1)(x-2)^2} dx = \int \left(\frac{2}{3(x+1)} + \frac{2}{3(x-2)} + \frac{7}{3(x-2)}\right) dx [/tex] which are all trivial integrals.

If those are correct values for A B and C it simplifies even more.
 
622
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your partial fractions is wrong. i get this system

A+B=0
-4A-B+C=1
4A-2B+C=0

then solve for a b and c again. then plug them in and replace your initial integral with your partial fractions. shouldn't be too bad from there. show your work if you're still stuck.
 
2,208
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Fine Gale, take the kill.
 
202
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Let me first check if I have it right.

To get partial fraction I would need 3 definitions, right?

Like

[tex]\frac{A}{x+1} + \frac{B}{x-2} + \frac{C}{(x-2)^2}[/tex].

Then solve for A,B and C.

then do the integration.

In my last attempt I ended up with a [tex]Ln[/tex] and a function.

Is it sposed to be like that?

Any help with the graph?
 
622
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you should end up with two Ln functions and a rational function after the integration. plus don't forget an integrating constant.

The graph hasn't been approved yet, so can't help you till we see it.
 
2,208
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Thats the correct equation, equate it to the original integrand and multiply out by the denominator, and solve the resulting system of equations for A B and C. Once you get those, itnegrate the equivalent partial fractions, each one will evaluate to a LN of a function except the last one.
 
202
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Thanx guys.
 

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