Question

How do you find `int (2x^2 -x -1) / ((x-1)(x^2 -4x)) dx` using partial fractions?

Answer 1

`int (2x^2-x-1)/((x-1)(x^2-4x)) dx = 9/4 ln abs(x-4) - 1/4 ln abs(x) + C`

Explanation:

`(2x^2-x-1)/((x-1)(x^2-4x)) = (color(red)(cancel(color(black)((x-1))))(2x+1))/(color(red)(cancel(color(black)((x-1))))(x^2-4x))`

`color(white)((2x^2-x-1)/((x-1)(x^2-4x))) = (2x+1)/(x(x-4))`

`color(white)((2x^2-x-1)/((x-1)(x^2-4x))) = a/(x-4) + b/x`

We can evaluate `a, b` using Heaviside's cover up method:

`a = (2(color(blue)(4))+1)/(color(blue)(4)) = 9/4`

`b = (2(color(blue)(0))+1)/((color(blue)(0))-4) = -1/4`

So:

`int (2x^2-x-1)/((x-1)(x^2-4x)) dx = int " "9/4(1/(x-4))-1/4(1/x) dx`

`color(white)(int (2x^2-x-1)/((x-1)(x^2-4x)) dx) = 9/4 ln abs(x-4) - 1/4 ln abs(x) + C`