How do you integrate #int (2x+1)/((x+3)(x-2)(x-3)) # using partial fractions?
1 Answer
Apr 19, 2016
Explanation:
First, focus on writing the fraction in terms of partial fractions:
#(2x+1)/((x+3)(x-2)(x-3))=A/(x+3)+B/(x-2)+C/(x-3)#
Multiply both sides by
#2x+1=A(x-2)(x-3)+B(x+3)(x-3)+C(x+3)(x-2)#
Expand:
#2x+1=A(x^2-5x+6)+B(x^2-9)+C(x^2+x-6)#
Sort by degree:
#2x+1=x^2(A+B+C)+x(-5A+C)+(6A-9B-6C)#
Giving the system:
#{(A+B+C=0),(-5A+C=2),(6A-9B-6C=1):}#
Solving which gives:
#{(A=-1/6),(B=-1),(C=7/6):}#
Thus, we know that
#int(2x+1)/((x+3)(x-2)(x-3))dx=int-1/(6(x+3))-1/(x-2)+7/(6(x-3))dx#
Splitting up the integral and refactoring constants:
#=-1/6int1/(x+3)dx-int1/(x-2)dx+7/6int1/(x-3)dx#
#=-1/6lnabs(x+3)-lnabs(x-2)+7/6lnabs(x-3)+C#