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

1 Answer
Sep 12, 2016

#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"#