How do you find int (x - 4 ) /((x-1)(x+2)(x-3)) dx∫x−4(x−1)(x+2)(x−3)dx using partial fractions?
1 Answer
int (x-4)/((x-1)(x+2)(x-3)) dx∫x−4(x−1)(x+2)(x−3)dx
= int 1/(2(x-1))-2/(5(x+2))-1/(10(x-3)) dx=∫12(x−1)−25(x+2)−110(x−3)dx
= 1/2 ln abs(x-1) - 2/5 ln abs(x+2) - 1/10 ln abs(x-3) + C=12ln|x−1|−25ln|x+2|−110ln|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)x−4(x−1)(x+2)(x−3)=Ax−1+Bx+2+Cx−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−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(x2−x−6)+B(x2−4x+3)+C(x2+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))=(A+B+C)x2+(−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
