How do you find #int (x - 4 ) /((x-1)(x+2)(x-3)) dx# using partial fractions?
1 Answer
#int (x-4)/((x-1)(x+2)(x-3)) dx#
#= int 1/(2(x-1))-2/(5(x+2))-1/(10(x-3)) dx#
#= 1/2 ln abs(x-1) - 2/5 ln abs(x+2) - 1/10 ln abs(x-3) + C#
Explanation:
Since the denominator is already factored into distinct linear factors, we are looking for a partial fraction decompositionof the form:
#(x-4)/((x-1)(x+2)(x-3)) = A/(x-1)+B/(x+2)+C/(x-3)#
#=(A(x+2)(x-3)+B(x-1)(x-3)+C(x-1)(x+2))/((x-1)(x+2)(x-3))#
#=(A(x^2-x-6)+B(x^2-4x+3)+C(x^2+x-2))/((x-1)(x+2)(x-3))#
#=((A+B+C)x^2+(-A-4B+C)x+(-6A+3B-2C))/((x-1)(x+2)(x-3))#
Equating coefficients, we get the system of simulataneous linear equations:
#{ (A+B+C = 0), (-A-4B+C=1), (-6A+3B-2C=-4) :}#
Add multiples of the first equation to the second and third equations to get:
#{ (A+B+C = 0), (-3B+2C=1), (9B+4C=-4) :}#
Add three times the second equation to the third equation to get:
#{ (A+B+C = 0), (-3B+2C=1), (10C=-1) :}#
Divide the third equation by
#{ (A+B+C = 0), (-3B+2C=1), (C=-1/10) :}#
Subtract multiples of the third equation from the other equations to get:
#{ (A+B = 1/10), (-3B=6/5), (C=-1/10) :}#
Divide the second equation by
#{ (A+B = 1/10), (B=-2/5), (C=-1/10) :}#
Subtract the second equation from the first to get:
#{ (A = 1/2), (B=-2/5), (C=-1/10) :}#
So:
#(x-4)/((x-1)(x+2)(x-3)) = 1/(2(x-1))-2/(5(x+2))-1/(10(x-3))#
So:
#int (x-4)/((x-1)(x+2)(x-3)) dx#
#= int 1/(2(x-1))-2/(5(x+2))-1/(10(x-3)) dx#
#= 1/2 ln abs(x-1) - 2/5 ln abs(x+2) - 1/10 ln abs(x-3) + C#
