Evaluate the integral? : # int_3^4 (x^2-3x+6)/(x(x-2)(x-1)) dx#

We first need to find the partial fraction decomposition of the integrand, which will be of the form;

# \ \ \ \ \ (x^2-3x+6)/(x(x-2)(x-1)) -= A/x + B/(x-2) + C/(x-1) #
# :. (x^2-3x+6)/(x(x-2)(x-1)) = (A(x-2)(x-1) + Bx(x-1) + Cx(x-2)) /(x(x-2)(x-1))#

And so:

# :. x^2-3x+6 = A(x-2)(x-1) + Bx(x-1) + Cx(x-2) #

We can easily find the constants #A,B,C# by setting #x# to the specific values that make the denominator 0 or by comparing coefficients. (With practice this can be done instantly using the "Cover Up" method).

Put #x=0 => 6=A(-2)(-1) \ \ \ \ \ \ \ \ => A=3#
Put #x=2 => 4-6+6=B(2)(1) \ \ \ \ \ => B=2#
Put #x=1 => 1-3+6=C(1)(-1) => \ C=-4#

So the integral can be written as:
# int_3^4 (x^2-3x+6)/(x(x-2)(x-1)) dx = int_3^4 3/x + 2/(x-2) -4/(x-1) dx#

Which we can now integrate as we know that

#d/dxln(ax+b)=a/(ax+b)#

Hence,

# int_3^4 (x^2-3x+6)/(x(x-2)(x-1)) dx = [3lnx+2ln|x-2| -4ln|x-1|]_3^4#
# int_3^4 (x^2-3x+6)/(x(x-2)(x-1)) dx = (3ln4+2ln2 -4ln3)-(3ln3+2ln1-4ln2)#
# int_3^4 (x^2-3x+6)/(x(x-2)(x-1)) dx = 3ln4+2ln2 -4ln3-3ln3-0+4ln2#
# int_3^4 (x^2-3x+6)/(x(x-2)(x-1)) dx = 3ln4+6ln2 -7ln3#
# int_3^4 (x^2-3x+6)/(x(x-2)(x-1)) dx = (2*3)ln2+6ln2 -7ln3#
# int_3^4 (x^2-3x+6)/(x(x-2)(x-1)) dx = 12ln2 -7ln3#
# int_3^4 (x^2-3x+6)/(x(x-2)(x-1)) dx ~~ 0.627480146042576#