unimatix
  • unimatix
Integral (3x+1)/(x^2-3x+2) dx
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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unimatix
  • unimatix
|dw:1441664655453:dw|
anonymous
  • anonymous
did you try partial fractions?
unimatix
  • unimatix
No hold I'll try that.

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unimatix
  • unimatix
Not sure how to do that in this case.
unimatix
  • unimatix
Maybe like a 1/3 x?
anonymous
  • anonymous
factor the denominator to \((x-1)(x-2)\) Then the original fraction can be written as \[\frac{ A }{ x-1 }+\frac{ B }{ x-2 }\]
anonymous
  • anonymous
\[\frac{ 3x+1 }{ (x-1)(x-2) }=\frac{ A }{ x-1 }+\frac{ B }{ x-2 }\]
unimatix
  • unimatix
Okay I got that.
unimatix
  • unimatix
I'm watching a video on partial fractions because it's been so long since I've seen them.
Jhannybean
  • Jhannybean
\[\frac{3x+1}{x^2-3x+2} = \frac{3x+1}{(x-2)(x-1)}\]\[\frac{3x+1}{(x-2)(x-1)} = \frac{A}{(x-2)}+\frac{B}{(x-1)}\]\[3x+1=A(x-1)+B(x-2)\]Yu could use a system of equations or the "zero" method to solve for the x's
anonymous
  • anonymous
add on the right then set thh denominators equal. \[A(x-1)+B(x-2)=3x+1\] Equate the coefficients and solve the system \[A+B=3\] \[-A-2B=1\]
anonymous
  • anonymous
And then once you have A and B the integral becomes \[\int\limits \left( \frac{ A }{ x-1 }+\frac{ B }{ x-2 } \right)dx\]
Jhannybean
  • Jhannybean
Let's say we took \(x=1\), then \[3(1)+1=A(1-1)+B(1-2)\]\[4=0-B\]\[B=-4\]\[3x+1=A(x-1)+B(x-2)\]\[3(2)+1=A(2-1)+0\]\[7=A\]
amistre64
  • amistre64
not sure if partials are needed, what is the derivative of the bottom to start with? can we modify the top?
Jhannybean
  • Jhannybean
Basically trying to find what makes A and B = 0... etc.
Jhannybean
  • Jhannybean
\[\int \left(\frac{7}{x-1}-\frac{4}{x-2}\right)dx\]... Trying to refresh my memory on Partial Fraction Decomposition as well. I found PFD easier because the polynomial at the bottom could be factored, and factoring leads me to partial fractions.. etc.
amistre64
  • amistre64
2x-3: 3x+1+(-x+4+x-4) 2x-3: 2x-3+(x+4) meh, would still need it for part of it i spose ...
Jhannybean
  • Jhannybean
\[7\int \left(\frac{1}{x-1}\right)dx -4\int \left(\frac{1}{x-2}\right)dx \] Im guessing?
Jhannybean
  • Jhannybean
@unimatix Are you capable of continuing from here?
unimatix
  • unimatix
I think so! Thanks for all your help everyone!
Jhannybean
  • Jhannybean
And can you show me how you would integrate this? Just want to make sure you do it right.

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