How do you integrate #int (x+5)/((x+3)(x-7)(x-5)) # using partial fractions?

from the given #int (x+5)/((x+3)(x-7)(x-5))# #dx# =

#intA/(x+3)##dx# + #B/(x-7)##dx# + #C/(x-5)#*#dx#

our equation becomes

#(x+5)/((x+3)(x-7)(x-5)#= #A/(x+3)+B/(x-7)+C/(x-5)#

it follows;

#(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))#

by using only the numerators;

#x+5=A(x^2-12x+35)+B(x^2-2x-15)+C(x^2-4x-21) #

collecting the like terms

#x+5=(A+B+C)x^2+(-12A-2B-4C)x+(35A-15B-21C)#

because the left side of the equation means

#0*x^2+1*x+5#

and the right side means

#(A+B+C)x^2+(-12A-2B-4C)x+(35A-15B-21C)#

the equations are now formed;

#A+B+C =0# first equation

#-12A-2B-4C=1# second equation

#35A-15B-21C=5# third equation

using your skills in solving 3 equations with 3 unknowns A, B, C

the values are #A=1/40#, #B=3/5#, and #C=-5/8#

Now, go back to the first line of the explanation to do the integration procedures.

#intA/(x+3)##dx# + #B/(x-7)##dx# + #C/(x-5)#*#dx#

#=int(1/40)/(x+3)##dx# + #int (3/5)/(x-7)##dx# + #int (-5/8)/(x-5)##dx#

final answer becomes

#int (x+5)/((x+3)(x-7)(x-5)# #dx#

#= 1/40*ln(x+3)+3/5*ln(x-7)-5/8*ln(x-5) + K#

for a definite integral, a constant K must be added.