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

Question

Christopher Allers · Accepted Answer

#-20/11ln|x-9|+6/7ln|x-5|-3/77ln|x+2|+const.#

We can rewrite the integrand expression in this way #(1-x^2)/((x-9)(x-5)(x+2))=A/(x-9)+B/(x-5)+C/(x+2)#

Making #x=0, 1 and -1# (it's convenient to make #x=1# and #x=-1# since then #1-x^2=0#), we get

#1/90=A/-9+B/-5+C/2# #0=A/-8+B/-4+C/3# #0=A/-10+B/-6+C/1#

Or #[[-1/9,-1/5,1/2],[-1/8,-1/4,1/3],[-1/10,-1/6,1]][[A],[B],[C]]=[[1/90],[0],[0]]#

Solving the system of variables #Delta =[[-1/9,-1/5,1/2],[-1/8,-1/4,1/3],[-1/10,-1/6,1]]=1/36+1/150+1/96-(1/80+1/162+1/40)=(1800+432+675-810-400-1620)/64800=77/64800# #Delta A=[[1/90,-1/5,1/2],[0,-1/4,1/3],[0,-1/6,1]]=-1/360-(-1/1620)=(-9+2)/3240=-7/3240# #Delta B=[[-1/9,1/90,1/2],[-1/8,0,1/3],[-1/10,0,1]]=-1/2700-(-1/720)=(-4+15)/10800=11/10800# #Delta C=[[-1/9,-1/5,1/90],[-1/8,-1/4,0],[-1/10,-1/6,0]]=1/4320-(1/3600)=(5-6)/21600=-1/21600#

So

#A=(Delta A)/Delta=(-7/3240)/(77/64800)=-1/11*20=-20/11# #B=(Delta B)/Delta=(11/10800)/(77/64800)=1/7*6=6/7# #C=(Delta C)/Delta=(-1/21600)/(77/64800)=-1/77*3=-3/77#

Subsequently, the initial phrase turns into

#=-20/11int dx/(x-9)+6/7int dx/(x-5)-3/77int dx/(x+2)# #=-20/11ln|x-9|+6/7ln|x-5|-3/77ln|x+2|+const.#

Cameron Alexander · Answer

undefined To integrate $ \int \frac{1-x^2}{(x-9)(x-5)(x+2)} $ using partial fractions, we first perform polynomial long division to ensure that the degree of the numerator is less than the degree of the denominator. After that, we express the rational function as a sum of partial fractions. Once we've found the appropriate form of the partial fraction decomposition, we can solve for the unknown coefficients by equating numerators and comparing coefficients. Finally, we integrate each term separately.

Since the degree of the numerator is less than the degree of the denominator, we can proceed with the partial fraction decomposition.

Let's express $ \frac{1-x^2}{(x-9)(x-5)(x+2)} $ as:

$$ \frac{1-x^2}{(x-9)(x-5)(x+2)} = \frac{A}{x-9} + \frac{B}{x-5} + \frac{C}{x+2} $$

Multiplying both sides by the denominator, we have:

$$ 1-x^2 = A(x-5)(x+2) + B(x-9)(x+2) + C(x-9)(x-5) $$

We'll now solve for $ A $, $ B $, and $ C $ by substituting appropriate values of $ x $ that make two of the terms zero.

After finding the values of $ A $, $ B $, and $ C $, we integrate each term separately, and then combine them to obtain the final result.