from the given integral, set up the equation using the variables A, B, and C
int (6-15x)/((x-1)(x+2)(x-4)) dx=int(A/(x-1)+B/(x+2)+C/(x-4))dx∫6−15x(x−1)(x+2)(x−4)dx=∫(Ax−1+Bx+2+Cx−4)dx
(6-15x)/((x-1)(x+2)(x-4))=A/(x-1)+B/(x+2)+C/(x-4)6−15x(x−1)(x+2)(x−4)=Ax−1+Bx+2+Cx−4
(6-15x)/((x-1)(x+2)(x-4))=(A(x^2-2x-8)+B(x^2-5x+4)+C(x^2+x-2))/((x-1)(x+2)(x-4))6−15x(x−1)(x+2)(x−4)=A(x2−2x−8)+B(x2−5x+4)+C(x2+x−2)(x−1)(x+2)(x−4)
(6-15x)/((x-1)(x+2)(x-4))6−15x(x−1)(x+2)(x−4)
=((A+B+C)x^2+(-2A-5B+C)x^1+(-8A+4B-2C)x^0)/((x-1)(x+2)(x-4))=(A+B+C)x2+(−2A−5B+C)x1+(−8A+4B−2C)x0(x−1)(x+2)(x−4)
The equations will be
A+B+C=0A+B+C=0
-2A-5B+C=-15−2A−5B+C=−15
-8A+4B-2C=6−8A+4B−2C=6
Simultaneous solution results to:
A=1A=1 and B=2B=2 and C=-3C=−3
final answer
color(red)(int (6-15x)/((x-1)(x+2)(x-4)) dx=)∫6−15x(x−1)(x+2)(x−4)dx=
color(red)(ln(x-1)+2ln(x+2)-3ln(x-4)+C_0)ln(x−1)+2ln(x+2)−3ln(x−4)+C0
God bless....I hope the explanation is useful.
