Calculus: Early Transcendentals, Chapter 7, 7.4, Section 7.4, Problem 12

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
:)