`int_0^1 (x-4)/(x^2 - 5x + 6) dx` Evaluate the integral

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Chapter 7, 7.4 - Problem 12 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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`int_0^1 (x-4)/(x^2 - 5x + 6) dx`

sol:

`int_0^1 (x-4)/(x^2 - 5x + 6) dx`

First let us solve the integral and then apply the limits later

so,

`int (x-4)/(x^2 - 5x + 6) dx`

=`int (x-4)/(x^2 - 3x -2x + 6) dx`

=`int (x-4)/(x(x - 3) -2(x -3)) dx`

=`int (x-4)/((x -2)(x -3)) dx`

Now by taking he partial fractions of `(x-4)/((x -2)(x -3)) `  we get

`(x-4)/((x -2)(x -3))` = `A/(x-2) +B/(x-3)`

                           = `(A(x-3)+B(x-2))/((x -2)(x -3))`  

so equating the numerators we get 

`x-4 ` = `A(x-3)+B(x-2)`

       = `x(A+B)+(-3A-2B)`

Now equating the co efficients of x and the constants we get

`A+B =1`

=> `A=1-B`

`-3A-2B =-4`

=>`3A+2B=4`

=> `3(1-B)+2B=4`

=> 3-3B+2B=4

=>`3-B=4`

=>` B= -1 so A = 1-B = 2`

Then 

`(x-4)/((x -2)(x -3)) ` = `A/(x-2) +B/(x-3)`

                            = `(2/(x-2)) - (1/(x-3))`

Now,

`int (x-4)/(x^2 - 5x + 6) dx ` =` int(x-4)/((x -2)(x -3)) dx`

                                        = `int ((2/(x-2)) - (1/(x-3))) dx`

=>`int (2/(x-2)) dx - int (1/(x-3)) dx`

=> `2ln(x-2) - ln(x-3) + c`

Now applying the limits 0 to 1 we get 

`int_0^1 (x-4)/(x^2 - 5x + 6) dx ` =` [2ln(x-2) - ln(x-3) ]_0^1`

                                          = `[2ln(1-2) - ln(1-3) ]-[2ln(0-2) - ln(0-3) ]`

       = `[2ln(-1) - ln(-2) ]-[2ln(-2) - ln(-3) ]`

        =`[ln(1^2) - ln(-2) ]-[ln((-2)^2) - ln(-3) ]`

      =`[ln(1) - ln(-2) ]-[ln(4) - ln(-3) ]`

     = `ln((-1)/2)) - ln((-4)/3))`

      =`ln(((-1)/2)/((-4)/3))`

     =` ln(3/4*2)`

      = `ln(3/8)`

is the solution 

:)

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