Question

How do you integrate `(x^2+x+1)/(1-x^2)` using partial fractions?

Answer 1

`int (x^2+x+1)/(1-x^2) dx = -x-3/2ln abs(x-1)+1/2 ln abs (x+1) + C`

Explanation:

`(x^2+x+1)/(1-x^2) = (-x^2-x-1)/(x^2-1)`

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

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

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

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

Use Heaviside's cover up method to find:

`A = (-(color(blue)(1))-2)/((color(blue)(1))+1) = -3/2`

`B = (-(color(blue)(-1))-2)/((color(blue)(-1))-1) = (-1)/(-2) = 1/2`

So:

`int (x^2+x+1)/(1-x^2) dx = int -1-3/2(1/(x-1))+1/2(1/(x+1)) dx`

`color(white)(int (x^2+x+1)/(1-x^2) dx) = -x-3/2ln abs(x-1)+1/2 ln abs (x+1) + C`