Integral (3x+1)/(x^2-3x+2) dx

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Integral (3x+1)/(x^2-3x+2) dx

Mathematics
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did you try partial fractions?
No hold I'll try that.

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Other answers:

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

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