Question

How do you integrate `(x+2)/(2x^3-8x)` using partial fractions?

Answer 1

`int \ (x+2)/(2x^3-8x) \ dx = 1/4ln|(x-2)/x| + C`

Explanation:

We seek:

`I = int \ (x+2)/(2x^3-8x) \ dx`

We can write as:

`I = int \ (x+2)/(2x(x^2-4)) \ dx`
`\ \ = 1/2 \ int \ (x+2)/(x(x+2)(x-2) \ dx` (removable discontinuity)
`\ \ = 1/2 \ int \ 1/(x(x-2) \ dx`

We can now decompose the integrand in to partial fractions:

`1/(x(x-2)) -= A/x + B/(x-2)`
`" " = (A(x-2)+Bx)/(x(x-2))`

Leading to the identity:

`1 -= A(x-2)+Bx`

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 = 0 => 1=-2A => A = -1/2`
Put `x = 2 => 1=2B => B = 1/2`

So we can now write:

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

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

`I = 1/4{ln|x-2| - ln|x| } + C`
`\ \ = 1/4ln|(x-2)/x| + C`