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