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

1 Answer
George C.
Oct 13, 2016

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

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