Question

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

Answer 1

`int_3^4 (x^2-3x+6)/(x(x-2)(x-1)) dx = 12ln2 -7ln3`

Explanation:

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`