Question

How do you integrate `int (4x^3 - 4x^2 - 16x + 7) / ((x + 1) (x -2))` using partial fractions?

Answer 1

`int \ (4x^3 - 4x^2 - 16x + 7) / ((x+1)(x-2)) \ dx = 2x^2 -5 ln |x+1| - 3ln|x-2| + C`

Explanation:

We seek:

`I = int \ (4x^3 - 4x^2 - 16x + 7) / ((x+1)(x-2)) \ dx`

First, we note that the order of the polynomial of the numerator is higher than that of the denominator. We therefore have the algebraic equivalent of a "top-heavy" fraction, and so we must first use algebraic long division:

# {: ( , , ul(+4x), ul(" "), ul(" "), ul(" "), ), ( x^2-x-2, ")", +4x^3, -4x^2, -16x, + 7, ), ( , , ul(+4x^3), ul(-4x^2), ul(-8x), ul(" "), -), ( , , , , -8x, +7, ) :} #

So, we find:

`(4x^3 - 4x^2 - 16x + 7) / (x^2-x-2) -= 4x + (-8x+7)/(x^2-x-2)`

We can now decompose the second term in to partial fractions:

`(-8x+7)/(x^2-x-2) -= (-8x+7)/((x+1)(x-2))`
`" " = A/(x+1) + B/(x-2)`
`" " = (A(x-2) + B(x+1)) / ((x+1)(x-2))`

Leading to the identity:

`-8x+7 = A(x-2) + B(x+1)`

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 = -1 => 8+7=A(-3) => A = -5`
Put `x = +2 => -16+7 = 3B => B = -3`

So we can now write:

`I = int \ 4x - 5/(x+1) - 3/(x-2) \ dx`

Which now consists of standard integral so we integrate to get:

`I = 2x^2 -5 ln |x+1| - 3ln|x-2| + C`