Question

How do you integrate `int (1-x^2)/((x-9)(x-5)(x+12))` using partial fractions?

Answer 1

`int (1-x^2)/((x-9)(x-5)(x+12)) dx`

`=-20/21 ln abs(x-9) + 6/17 ln abs(x-5) - 143/357 ln abs(x+12) + "constant"`

Explanation:

Let's look at the general problem:

`int (ax^2+bx+c)/((x-d)(x-e)(x-f)) dx`

where `d, e, f` are distinct.

We can split the integrand into partial fractions like this:

`(ax^2+bx+c)/((x-d)(x-e)(x-f))=A/(x-d)+B/(x-e)+C/(x-f)`

where `A, B, C` can be determined using Heaviside's cover up method:

`A = (acolor(blue)(d)^2+bcolor(blue)(d)+c)/((color(blue)(d)-e)(color(blue)(d)-f))`

`B = (acolor(blue)(e)^2+bcolor(blue)(e)+c)/((color(blue)(e)-d)(color(blue)(e)-f))`

`C = (acolor(blue)(f)^2+bcolor(blue)(f)+c)/((color(blue)(f)-d)(color(blue)(f)-e))`

Then:

`int (ax^2+bx+c)/((x-d)(x-e)(x-f)) dx`

`=int (A/(x-d)+B/(x-e)+C/(x-f)) dx`

`=A ln abs(x-d) + B ln abs(x-e) + C ln abs(x-f) + "constant"`

In our example:

`a=-1`, `b = 0`, `c = 1`, `d = 9`, `e = 5`, `f = -12`

So:

`A=(-80)/((4)(21)) = -20/21`

`B=(-24)/((-4)(17)) = 6/17`

`C=(-143)/((-21)(-17)) = -143/357`

So

`int (1-x^2)/((x-9)(x-5)(x+12)) dx`

`=-20/21 ln abs(x-9) + 6/17 ln abs(x-5) - 143/357 ln abs(x+12) + "constant"`