Since all the factors in the denominator of (x+1)/((x-9)(x+3)(x-2))x+1(x−9)(x+3)(x−2) are linear (yay!), our partial fraction decomposition will be:
A/(x-9)+B/(x+3)+C/(x-2)Ax−9+Bx+3+Cx−2
For now, let's ignore the integral and focus on decomposing this fraction:
(x+1)/((x-9)(x+3)(x-2))=A/(x-9)+B/(x+3)+C/(x-2)x+1(x−9)(x+3)(x−2)=Ax−9+Bx+3+Cx−2
x+1=A(x+3)(x-2)+B(x-9)(x-2)+C(x-9)(x+3)x+1=A(x+3)(x−2)+B(x−9)(x−2)+C(x−9)(x+3)
Let x=9x=9 to find the value of AA:
9+1=A(9+3)(9-2)+B(9-9)(9-2)+C(9-9)(9+3)9+1=A(9+3)(9−2)+B(9−9)(9−2)+C(9−9)(9+3)
10=84A10=84A
A=10/84=5/42A=1084=542
Let x=-3x=−3 to find the value of BB:
-3+1=A(-3+3)(-3-2)+B(-3-9)(-3-2)+C(-3-9)(-3+3)−3+1=A(−3+3)(−3−2)+B(−3−9)(−3−2)+C(−3−9)(−3+3)
-2=60B−2=60B
B=-2/60=-1/30B=−260=−130
And let x=2x=2 to find the value of CC:
2+1=A(2+3)(2-2)+B(2-9)(2-2)+C(2-9)(2+3)2+1=A(2+3)(2−2)+B(2−9)(2−2)+C(2−9)(2+3)
3=-35C3=−35C
C=-3/35C=−335
Therefore, (x+1)/((x-9)(x+3)(x-2))=(5/42)/(x-9)-(1/30)/(x+3)-(3/35)/(x-2)x+1(x−9)(x+3)(x−2)=542x−9−130x+3−335x−2. Our integral is now:
int(5/42)/(x-9)-(1/30)/(x+3)-(3/35)/(x-2)dx∫542x−9−130x+3−335x−2dx
Using the sum rule, this becomes:
int(5/42)/(x-9)dx-int(1/30)/(x+3)dx-int(3/35)/(x-2)dx∫542x−9dx−∫130x+3dx−∫335x−2dx
Integrating these gives:
5/42int1/(x-9)dx-1/30int1/(x+3)dx-3/35int1/(x-2)dx542∫1x−9dx−130∫1x+3dx−335∫1x−2dx
5/42lnabs(x-9)-1/30lnabs(x+3)-3/35lnabs(x-2)+C542ln|x−9|−130ln|x+3|−335ln|x−2|+C
