Question

How do I evaluate this integral?

`int_0^1(x^3-4x-9)/(x^2-x-6)dx`

Answer 1

We seek:

`I = int_0^1 \ (x^3-4x-9)/(x^2-x-6) \ dx`

As the degree of the numerator is higher than that of the denominator, then we can use algebraic long division to reduce the ordre of the fraction. We find that:

`(x^3-4x-9) divide (x^2-x-6) = (x+1)` with a remainder `(3x-3)`

Thus:

`I = int_0^1 \ (x+1) + (3x-3)/(x^2-x-6) \ dx`

We now decompose the second component into partial fractions:

`(3x-3)/(x^2-x-6) -= (3x-3)/((x+2)(x-3))`
`" " = A/(x+2) + B/(x-3)`
`" " = ( A(x-3) + B(x+2)) /((x+2)(x-3))`

Leading to:

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

Where `A,B` are constants that are to be determined. We can find them by substitutions (In practice we do this via the "cover up" method:

Put `x = -2 => -6-3 = A(-2-3) => A = 9/5`
Put `x = +3 => 9-3=B(3+2) => B = 6/5`

Using this we can write the integral as:

`I = int_0^1 \ (x+1) + (9/5)/(x+2) + (6/5)/(x-3) \ dx`

Which we can now readily integrate to give:

`I = [1/2x^2+x+9/5ln|x+2| + 6/5ln|x-3|]_0^1`
`\ \ = (1/2+1+9/5ln3+6/5ln2) - (0+0+9/5ln2+6/5ln3)`

`\ \ = 3/2 + 9/5ln3+6/5ln2 - 9/5ln2-6/5ln3`

`\ \ = 3/2 + (9/5-6/5)ln3 + (6/5-9/5)ln2`

`\ \ = 3/2 + 3/5ln3 -3/5ln2`

`\ \ = 3/2 + 3/5(ln3 -ln2)`

`\ \ = 3/2 + 3/5ln(3/2)`