from the given int (x+5)/((x+3)(x-7)(x-5))∫x+5(x+3)(x−7)(x−5) dxdx =
intA/(x+3)∫Ax+3dxdx + B/(x-7)Bx−7dxdx + C/(x-5)Cx−5*dxdx
our equation becomes
(x+5)/((x+3)(x-7)(x-5)x+5(x+3)(x−7)(x−5)= A/(x+3)+B/(x-7)+C/(x-5)Ax+3+Bx−7+Cx−5
it follows;
(x+5)/((x+3)(x-7)(x-5)x+5(x+3)(x−7)(x−5)=
(A(x-7)(x-5)+B(x+3)(x-5)+C(x+3)(x-5))/((x+3)(x-5)(x-7))A(x−7)(x−5)+B(x+3)(x−5)+C(x+3)(x−5)(x+3)(x−5)(x−7)
by using only the numerators;
x+5=A(x^2-12x+35)+B(x^2-2x-15)+C(x^2-4x-21)
x+5=A(x2−12x+35)+B(x2−2x−15)+C(x2−4x−21)
collecting the like terms
x+5=(A+B+C)x^2+(-12A-2B-4C)x+(35A-15B-21C)x+5=(A+B+C)x2+(−12A−2B−4C)x+(35A−15B−21C)
because the left side of the equation means
0*x^2+1*x+50⋅x2+1⋅x+5
and the right side means
(A+B+C)x^2+(-12A-2B-4C)x+(35A-15B-21C)(A+B+C)x2+(−12A−2B−4C)x+(35A−15B−21C)
the equations are now formed;
A+B+C =0A+B+C=0 first equation
-12A-2B-4C=1−12A−2B−4C=1 second equation
35A-15B-21C=535A−15B−21C=5 third equation
using your skills in solving 3 equations with 3 unknowns A, B, C
the values are A=1/40A=140, B=3/5B=35, and C=-5/8C=−58
Now, go back to the first line of the explanation to do the integration procedures.
intA/(x+3)∫Ax+3dxdx + B/(x-7)Bx−7dxdx + C/(x-5)Cx−5*dxdx
=int(1/40)/(x+3)=∫140x+3dxdx + int (3/5)/(x-7)∫35x−7dxdx + int (-5/8)/(x-5)∫−58x−5dxdx
final answer becomes
int (x+5)/((x+3)(x-7)(x-5)∫x+5(x+3)(x−7)(x−5) dxdx
= 1/40*ln(x+3)+3/5*ln(x-7)-5/8*ln(x-5) + K=140⋅ln(x+3)+35⋅ln(x−7)−58⋅ln(x−5)+K
for a definite integral, a constant K must be added.
