How do you integrate #int (1-x^2)/((x-9)(x+6)(x-2)) # using partial fractions?
↳Redirected from
"Which following pairs of atoms, have a lower electron affinity? a) Ca,K b) I,F c) Li, Ra. I seriously don't know anything about electron affinity all ik that it can buy another element"
#int(1-x^2)/((x-9)(x+6)(x-2))dx#
#=-16/21lnabs(x-9)-7/24lnabs(x+6)+3/56lnabs(x-2)+C#
Conveniently, the denominator in this problem is already factored.
To apply partial fraction decomposition, we want to express
#int(1-x^2)/((x-9)(x+6)(x-2))dx#
in the form
#int(A/(x-9)+B/(x+6)+C/(x-2))dx#
So we can equate
#(1-x^2)/((x-9)(x+6)(x-2))=A/(x-9)+B/(x+6)+C/(x-2)#
Multiplying both sides by the denominator of the left, we get:
#1-x^2=A(x+6)(x-2)+B(x-9)(x-2)+C(x-9)(x+6)#
Now let's expand the parentheses and group similar terms:
#1-x^2=A(x^2+4x-12)+B(x^2-11x+18)+C(x^2-3x-54)#
#1-x^2=(A+B+C)x^2+(4A-11B-3C)x+(-12A+18B-54C)#
Now we can compare the coefficients on both sides, and write a system of three equations:
#A+B+C=-1#
#4A-11B-3C=0#
#-12A+18B-54C=1#
Solving this system (whichever way you choose) gives:
#A=-16/21#
#B=-7/24#
#C=3/56#
So now we can rewrite the integral as:
#int(-16/21(1)/(x-9)-7/24(1)/(x+6)+3/56(1)/(x-2))dx#
#rArr-16/21int1/(x-9)dx-7/24int1/(x+6)dx+3/56int1/(x-2)dx#
Integrating, we get:
#-16/21lnabs(x-9)-7/24lnabs(x+6)+3/56lnabs(x-2)+C#
