Integration by Partial Fractions

  • Thread starter cathy
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  • #1
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Homework Statement


(2x^3-2x+1)/(x^2/3x)
Find the integral.





2. The attempt at a solution

So I've been on this problem for like an hour now and I don't know what I'm doing wrong. So I used long division and got
2x+ (4x+1)/(x^2-3x)

∫2x + ∫(4x+1)/(x^2-3x)
= x^2 + ∫(4x+1)/(x^2-3x).
For that next part, I used partial fractions.
A/x + B/(x-3) = 4x+1
Solved for a= -1/3
Solved for b= 13/3

Put everything together and solving for that integral, I got

x^2- 1/3ln(x) + 13/3ln(x-3)

and it's wrong.
Where am I going wrong? I'm so confused.
Thank for any help in advance.
 

Answers and Replies

  • #2
vanhees71
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Obviously there went something wrong with your partial-fraction calculation. I'd just do the following
[tex]\frac{2x^3-2x+1}{x^2-3x}=\frac{2x(x^2-3x)+6(x^2-3x)+16x+1}{x^2-3x}=2x+6 + \frac{16x+1}{x^2-3x}.[/tex]
The first two terms are easily integrated. For the 2nd one you can find the partial fractions with the usual ansatz easily.
 
  • #3
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But i was told to use long division first, which was why I started with 2x+ (4x+1)/(x^2-3x).
And from there, I am confused, as to how I got the answer part wrong.
 
  • #4
vanhees71
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What I did is nothing else then long division (written in a way I can achieve with LaTeX).
 
  • #5
HallsofIvy
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What vanhees71 is trying to tell you is that your long division is wrong. You should have "2x+ 6" as the quotient, not just "2x". And the remainder is "16x+ 1", not "4x+ 1".

I suspect you did not write the "[itex]0x^2[/itex]" in the dividend and then mistook the "[itex]-6x^2[/itex]" in the division for "[itex]-6x[/itex]".
 
  • #6
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Homework Statement


(2x^3-2x+1)/(x^2/3x)
Find the integral.
The above has a typo. I'm pretty sure you meant the integrand to be (2x3 - 2x + 1)/(x2 - 3x)
2. The attempt at a solution

So I've been on this problem for like an hour now and I don't know what I'm doing wrong. So I used long division and got
2x+ (4x+1)/(x^2-3x)

∫2x + ∫(4x+1)/(x^2-3x)
= x^2 + ∫(4x+1)/(x^2-3x).
For that next part, I used partial fractions.
A/x + B/(x-3) = 4x+1
Solved for a= -1/3
Solved for b= 13/3

Put everything together and solving for that integral, I got

x^2- 1/3ln(x) + 13/3ln(x-3)

and it's wrong.
Where am I going wrong? I'm so confused.
Thank for any help in advance.
 
  • #7
90
0
Ohh, such a simple mistake with the long division. I was forgetting to write the variables I didn't use with 0. Thank you so much everyone.
 

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